Cantor Confusion
the_...@yahoo.com
seems that people are confused or uncomfortable with it, so
I've tried to summarize it to the simplest terms:
1. Assume there is a list containing all the reals.
2. Show that a real can be defined/constructed from that list.
3. Show why the real from step 2 is not on the list.
4. Conclude that the premise is wrong because of the contradiction.
The steps are simple except for a possible debate about defined /
constructed. I don't think anyone believes the proof is invalid
because of that debate however.
There seems to be another area that seems to be a problem
though. The problem is that step #2 doesn't seem valid. If we
assume the list contains all the real numbers, then defining or
constructing a real number in terms of that list would be
self-referential. The number from step #2, that is normally defined
digit-by-digit along the diagonal, must have its digits (or at least
one of them) defined as not equal to itself, if we are to assume the
list contains all the real numbers. Certainly the conclusion in that
case is that the premise is wrong or that the construction is not
valid, but the conclusion can't be simply that the premise is wrong.
This same problem appears in the "power-set" theorem, where we
have a definition of a set, S, which is a subset of N, defined in
terms of the image of a function, f, whose image is assumed to be
the power-set of N. If the image of f is assumed to be the power-set
of N and S is defined in terms of f, then S must necessarily be
defined in terms of itself. Again, if we assume that the image of
f is P(N), then defining S as a set whose elements are defined
to be elements not in the image of f is a self-referential definition
of S because S is also a subset of N, making it meaningless.
Certainly a meaningless definition can't be used to prove a
contradiction.
I'm guessing that the "discussions" that occur stem from the fact
that mathematicians disagree that the seemingly self-referential
definitions are a problem but it's not intuitively obvious why that is
so, therefore many people feel the need to try to refute the proofs.
The problem is that it really isn't clear why mathematicians seem to
accept the self-referential definitions.
cbr...@cbrownsystems.com
> Cantor's proof is one of the most popular topics on this NG. It
> seems that people are confused or uncomfortable with it, so
> I've tried to summarize it to the simplest terms:
>
> 1. Assume there is a list containing all the reals.
> 2. Show that a real can be defined/constructed from that list.
> 3. Show why the real from step 2 is not on the list.
> 4. Conclude that the premise is wrong because of the contradiction.
>
> The steps are simple except for a possible debate about defined /
> constructed. I don't think anyone believes the proof is invalid
> because of that debate however.
>
> There seems to be another area that seems to be a problem
> though. The problem is that step #2 doesn't seem valid. If we
> assume the list contains all the real numbers, then defining or
> constructing a real number in terms of that list would be
> self-referential. The number from step #2, that is normally defined
> digit-by-digit along the diagonal, must have its digits (or at least
> one of them) defined as not equal to itself, if we are to assume the
> list contains all the real numbers. Certainly the conclusion in that
> case is that the premise is wrong or that the construction is not
> valid, but the conclusion can't be simply that the premise is wrong.
>
Why not?
The thing is, we can "see" that the argument itself (i.e., that we can
construct a number distinct from the list of numbers which we assume is
complete) follows logically, in the same way that we "see" that the
premises "All men are mortal" and "Socrates is a man" implies that
"Socrates is mortal".
In the given case, if we accept the premises, the result of the
argument is that if the premise is true ("there is a list of all
reals"), then it still follows that "every list of reals, complete or
not, misses at least one real", and thus the absurd conclusion ("the
complete list of all reals misses at least one real").
To continue the analogy, if we accept the premises "All men are
mortal", "Socrates is a man", and "Socates is not mortal", then it
follows that at least one of our /premises/ is false, not that the
logic which demonstrates the absurdity is false.
> This same problem appears in the "power-set" theorem, where we
> have a definition of a set, S, which is a subset of N, defined in
> terms of the image of a function, f, whose image is assumed to be
> the power-set of N. If the image of f is assumed to be the power-set
> of N and S is defined in terms of f, then S must necessarily be
> defined in terms of itself. Again, if we assume that the image of
> f is P(N), then defining S as a set whose elements are defined
> to be elements not in the image of f is a self-referential definition
> of S because S is also a subset of N, making it meaningless.
> Certainly a meaningless definition can't be used to prove a
> contradiction.
A definition can be meaningful, and yet not be satisfied by any
mathmeatical object. For example, suppose we say that a natural number
p is "oddven" is p is both odd and even. This definition is meaningful
(in the sense that, for any natural number n, we can unambiguously say
that n either is or is not oddven), but there are no natural numbers
which are oddven.
In the same way, the definition of "T is a list of all real numbers" is
meaningful, and yet there is no such list T.
>
> I'm guessing that the "discussions" that occur stem from the fact
> that mathematicians disagree that the seemingly self-referential
> definitions are a problem but it's not intuitively obvious why that is
> so, therefore many people feel the need to try to refute the proofs.
> The problem is that it really isn't clear why mathematicians seem to
> accept the self-referential definitions.
Because they're not "self-referential" in the way you seem to use the
term. The proof that there is no T such that T is a complete list of
the reals follows from the fact that every list of reals misses at
least one real, just as the proof that there are no natural numbers n
which are oddven follows from the fact that a number which is even must
neccessarily not be odd.
Cheers - Chas
Peter Webb
<the_...@yahoo.com> wrote in message
news:1159410937.0...@h48g2000cwc.googlegroups.com...
Its not "self-referential" in the sense of being a circular argument.
You produce ANY list of all Reals, I can show you a missing real. Therefore
I can do it for ALL lists, and hence there is no complete list of Reals.
Obviously you have to tell me the list first, as there is no single Real
which is missing from every list, or that Real could be made the first entry
on a new list. The Real is different on every list, and depends on the list.
You have three balls, red green and blue. I state you cannot put them all
into two boxes, such that no box contains two or more balls. Here is my
proof:
If you put a red ball into one box, green into another box, then there is
nowhere to put the blue ball (and so on for the other two possibilities).
This requires me to know which boxes you put each ball into, but its not
self-referential. If you the blue ball is left out, you can't just put the
blue ball into box #1 and move the red ball out and say - "see, you can put
all the balls into two boxes". This is what anti-Cantor cranks do all the
time - we identify the Real not on the list (the ball that has been
excluded) and then they change the list to include that Real (change what
boxes the other balls go into) and say "see, it is on the list" ("see, I can
put the blue ball into a box").
Virgil
cbr...@cbrownsystems.com wrote:
> the_...@yahoo.com wrote:
> > Cantor's proof is one of the most popular topics on this NG. It
> > seems that people are confused or uncomfortable with it, so
> > I've tried to summarize it to the simplest terms:
> >
> > 1. Assume there is a list containing all the reals.
> > 2. Show that a real can be defined/constructed from that list.
> > 3. Show why the real from step 2 is not on the list.
> > 4. Conclude that the premise is wrong because of the contradiction.
That is a proof by contradiction, which many constructionists object to.
One can modify it slightly to get a more direct proof:
1. Assume one is given any list of reals (i.e., an arbitrary function
with domain N and codomain R, the only condition being that there is one
real for each natural number)
2. Show that a real can be defined/constructed from that list in such a
way as not to be a member of that list.
3. Conclude that every list must omit at least one real, so no list is
complete.
muec...@rz.fh-augsburg.de
Peter Webb schrieb:
> You produce ANY list of all Reals, I can show you a missing real. Therefore
> I can do it for ALL lists, and hence there is no complete list of Reals.
>
> Obviously you have to tell me the list first, as there is no single Real
> which is missing from every list,
No?
But the set of all lists is countable (as is any quantized or
discontinuous set), so is the set of all list entries. Nevertheless,
there is no real number missing in every list? So every real number is
in at least one of the list? So every real number is one element of a
countable set of entries? And there is nothing real really outside of
this countable set?
Regards, WM
Dave Seaman
> Peter Webb schrieb:
>> You produce ANY list of all Reals, I can show you a missing real. Therefore
>> I can do it for ALL lists, and hence there is no complete list of Reals.
>>
>> Obviously you have to tell me the list first, as there is no single Real
>> which is missing from every list,
> No?
> But the set of all lists is countable (as is any quantized or
> discontinuous set),
Wrong.
> so is the set of all list entries. Nevertheless,
> there is no real number missing in every list? So every real number is
> in at least one of the list? So every real number is one element of a
> countable set of entries? And there is nothing real really outside of
> this countable set?
> Regards, WM
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Arturo Magidin
<the_...@yahoo.com> wrote:
>Cantor's proof is one of the most popular topics on this NG. It
>seems that people are confused or uncomfortable with it, so
>I've tried to summarize it to the simplest terms:
>
>1. Assume there is a list containing all the reals.
>2. Show that a real can be defined/constructed from that list.
>3. Show why the real from step 2 is not on the list.
>4. Conclude that the premise is wrong because of the contradiction.
This is hardly the simplest terms. Much simpler is to do a ->direct<-
proof instead of a proof by contradiction.
1. Take ANY list of real numbers.
2. Show that a real can be defined/constructed from that list.
3. Show that the real from step 2 is not on the list.
4. Conclude that no list can contain all reals.
Why insist on proof by contradiction? It just begs the other person to
misidentify what is "the" premise that is false. Maybe the constructed
number is not really constructed? Maybe the number is not really a
real? Etc.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Randy Poe
muec...@rz.fh-augsburg.de wrote:
> Peter Webb schrieb:
>
> > You produce ANY list of all Reals, I can show you a missing real. Therefore
> > I can do it for ALL lists, and hence there is no complete list of Reals.
> >
> > Obviously you have to tell me the list first, as there is no single Real
> > which is missing from every list,
>
> No?
No. Let x be any real. Then I can certainly create a list
{a_1, a_2, ...} with a_1 = x.
> But the set of all lists is countable
No it isn't.
- Randy
Ross A. Finlayson
Hi.
There are a variety of actively researched set theories that have
representations of numbers in them where the powerset or antidiagonal
argument is not said to hold. There are a variety of most types of
applied mathematics that ignore transfinite cardinals.
Cantor's (Georg Cantor, mathematician's) expectation that there is a
universe, is called here the domain principle. In ZF, which is taught
to a lot of people, there is no universe. So if you use a universe in
a set theory, then it is not ZF.
Remove all the (non-logical) axioms from any theory and then it is the
null axiom theory. If you're interested in a theory that is designed
with the goals of being consistent and complete, I've written some
thousands of pages about it to sci.math.
Ah, I had some excellent thoughts about definitions the other day, very
comforting, in the context as above.
Did you know some mathematicians divide by zero?
The universe is infinite, infinite sets are equivalent.
Ross
William Hughes
muec...@rz.fh-augsburg.de wrote:
> Peter Webb schrieb:
>
> > You produce ANY list of all Reals, I can show you a missing real. Therefore
> > I can do it for ALL lists, and hence there is no complete list of Reals.
> >
> > Obviously you have to tell me the list first, as there is no single Real
> > which is missing from every list,
>
> No?
>
> But the set of all lists is countable
Twaddle.
Let A ={ (x,0,0,0,...) | x is a real number }
Then A is a set of lists and A has as many elements
as there are real numbers.
>(as is any quantized or
> discontinuous set), so is the set of all list entries. Nevertheless,
> there is no real number missing in every list? So every real number is
> in at least one of the list? So every real number is one element of a
> countable set of entries? And there is nothing real really outside of
> this countable set?
>
Note any countable set of entries is a list (possibly
with a different ordering).
This is staight quantifier dyslexia
For every real number x, there exists a list, L_x such
that x is a member of L_x
There exists a list L, such that every real number x is
a member of L.
The first is true, the second is false. There is no way to put
all the L_x together to get a "countable set of entries"
(the list L).
- William Hughes
guenther vonKnakspot
drown your sorry feeble mind in beer at the Oktoberfest? Maybe they
will beat the crap out of you for good there.
Dave L. Renfro
> But the set of all lists is countable (as is any quantized
> or discontinuous set), so is the set of all list entries.
> Nevertheless, there is no real number missing in every
> list? So every real number is in at least one of the
> list? So every real number is one element of a countable
> set of entries? And there is nothing real really outside
> of this countable set?
The set of all lists of _what_? Also, "lists" is a term
that for some people means one thing, and for other people
it can mean something else. If you're going to argue about
details, rather than give a general overview of what's
going on, you should use the correct mathematical terms,
such as one-to-one correspondences (i.e. bijective functions).
Otherwise, the discussion is going to degenerate into
disagreements over word usage. In fact, you're doing it
yourself by bringing up "quantized" and "discontinuous",
two words that, to most people in math, have no direct
relevance to the discussion at hand.
It never ceases to amaze me that people can have so much
trouble with the main gist (not the details) of the
argument. It's the same way that people argue in so many
other situations, if not as rigorously. If your boss says
you have to do such and such, and you want to explain why
it's not possible, you say, well, if I did such and such,
then ... [insert consequence boss would not wish to allow].
If you want to prove someone is innocent of a crime in a
court of law, it's often enough to prove that the defendant
was somewhere else at the time (the contradiction being
that one person can't be in two places at the same time).
If you want to prove that someone isn't your child,
a negative DNA match will do it (the contradiction being
that a sufficiently strong DNA mismatch cannot occur
in a parent/offspring pair).
If you want to prove the set of real numbers is uncountable,
one way is to show that it's impossible for the set of real
numbers to be countable. I realize there are some issues
related to the law of excluded middle, but it seems to me
that the confusion people have with the diagonal argument
is more rooted in the details than with the overall issue
of the law of excluded middle. Anyway, as has been said
over and over again in here, the diagonal argument for
the uncountability of the reals doesn't have to be
an indirect argument. One can simply state and prove
the result in the following form: Each sequence of real
numbers omits at least one real number. (Sequence being
a 1-1 and onto function from the positive integers to
the real numbers.)
Dave L. Renfro
jmor...@idirect.com
the_...@yahoo.com wrote:
> Cantor's proof is one of the most popular topics on this NG. It
> seems that people are confused or uncomfortable with it, so
> I've tried to summarize it to the simplest terms:
>
> 1. Assume there is a list containing all the reals.
> 2. Show that a real can be defined/constructed from that list.
> 3. Show why the real from step 2 is not on the list.
> 4. Conclude that the premise is wrong because of the contradiction.
Isn't it important to ASSUME that the original list (in #1 above) is in
a one-to-one correspondence to the integers. You need this to count
down and across to construct a real that CAN'T be one of these in the
list. You need a bigger infinity...
Similarly, if you make a horizontal and vertical list of all integers,
going to infinity in both directions, then ANY rational can be placed
uniquely on that grid, and assigned a unique integer position number.
You DON'T need a bigger infinity here...
Arturo Magidin
jmor...@idirect.com <jmor...@idirect.com> wrote:
>
>the_...@yahoo.com wrote:
>> Cantor's proof is one of the most popular topics on this NG. It
>> seems that people are confused or uncomfortable with it, so
>> I've tried to summarize it to the simplest terms:
>>
>> 1. Assume there is a list containing all the reals.
>> 2. Show that a real can be defined/constructed from that list.
>> 3. Show why the real from step 2 is not on the list.
>> 4. Conclude that the premise is wrong because of the contradiction.
>
>
>Isn't it important to ASSUME that the original list (in #1 above) is in
>a one-to-one correspondence to the integers.
I assume you mean "positive integers"... (obviously, it is also true
if you use all integers, but the OP is trying to simplify...)
> You need this to count
>down and across to construct a real that CAN'T be one of these in the
>list. You need a bigger infinity...
Hmmm... A "list" in this context is understood to be a function from
the natural number (or from the positive integers) into the reals. The
proof does not require the function to be one-to-one, but it ->does<-
require it to be a ->function<- (defined at all inputs).
As noted elsewhere, the argument does NOT really require the
assumption that the list is complete (i.e., that the map is
->surjective<-). Even assuming that, you do not need the map to be
injective in order for steps 2 and 3 to work. You just need the map to
be defined at every n>0.
Dave Seaman
> the_...@yahoo.com wrote:
>> Cantor's proof is one of the most popular topics on this NG. It
>> seems that people are confused or uncomfortable with it, so
>> I've tried to summarize it to the simplest terms:
>>
>> 1. Assume there is a list containing all the reals.
>> 2. Show that a real can be defined/constructed from that list.
>> 3. Show why the real from step 2 is not on the list.
>> 4. Conclude that the premise is wrong because of the contradiction.
> Isn't it important to ASSUME that the original list (in #1 above) is in
> a one-to-one correspondence to the integers. You need this to count
> down and across to construct a real that CAN'T be one of these in the
> list. You need a bigger infinity...
No, you don't need any such assumption in order to apply the diagonal
argument. The construction of a real depends only on the existence of a
mapping f: N -> R. It is not necessary to assume that f is injective,
surjective, or bijective. For each such f, the proof yields a real
number not in the range of f. Hence, it follows that no such f can be
surjective.
> Similarly, if you make a horizontal and vertical list of all integers,
> going to infinity in both directions, then ANY rational can be placed
> uniquely on that grid, and assigned a unique integer position number.
> You DON'T need a bigger infinity here...
You don't need a bigger infinity to apply the diagonal argument. You
are putting the cart before the horse. It is a *consequence* (not a
prerequisite) of the diagonal argument that |R| > |N|.
Virgil
muec...@rz.fh-augsburg.de wrote:
That presumes, falsely, that the set of lists is itself countable,
david petry
the_...@yahoo.com wrote:
> Cantor's proof is one of the most popular topics on this NG. It
> seems that people are confused or uncomfortable with it, so
I don't know if anyone in this newsgroup is confused by Cantor's proof,
but what you have written misses the whole point of the endless debate
that takes place here.
The argument of the "anti-Cantorians" is that "real" mathematics is
computationally testable. That is, valid, meaningful mathematical
statements must make predictions about the outcome of computational
experiments. (And note that any mathematics that is potentially
applicable must satisfy that requirement) And hence, all objects that
exist in the world of "real" mathematics must be computable. And there
cannot be more than countably many such objects.
If you start with a well defined list of well defined real numbers (so
that every digit of every real number can be computed), then the
diagonal method gives us a way of constructing a new real number not on
that list, but that certainly does not imply that the well defined real
numbers are uncountable in Cantor's intended sense of the word.
Aatu Koskensilta
> In article <1159417542.4...@i3g2000cwc.googlegroups.com>,
> cbr...@cbrownsystems.com wrote:
>> the_...@yahoo.com wrote:
>>> 1. Assume there is a list containing all the reals.
>>> 2. Show that a real can be defined/constructed from that list.
>>> 3. Show why the real from step 2 is not on the list.
>>> 4. Conclude that the premise is wrong because of the contradiction.
>
> That is a proof by contradiction, which many constructionists object to.
No, they don't, presuming you mean constructivists. A direct
constructivistic proof of ~A is a proof of contradiction from A. What
one can't do constructively is to prove A by proving that ~A leads to a
contradiction.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Virgil
Aatu Koskensilta <aatu.kos...@xortec.fi> wrote:
> Virgil wrote:
> > In article <1159417542.4...@i3g2000cwc.googlegroups.com>,
> > cbr...@cbrownsystems.com wrote:
> >> the_...@yahoo.com wrote:
> >>> 1. Assume there is a list containing all the reals.
> >>> 2. Show that a real can be defined/constructed from that list.
> >>> 3. Show why the real from step 2 is not on the list.
> >>> 4. Conclude that the premise is wrong because of the contradiction.
> >
> > That is a proof by contradiction, which many constructionists object to.
>
> No, they don't, presuming you mean constructivists. A direct
> constructivistic proof of ~A is a proof of contradiction from A. What
> one can't do constructively is to prove A by proving that ~A leads to a
> contradiction.
In my understanding, a "proof by contradiction" is essentially a proof
of A by proving ~A false, which requires accepting the law of the
excluded middle.
I understood that constructivists did not much care for the law of the
excluded middle.
the_...@yahoo.com
that avoids the problem of the self-reference is to consider ANY
list of real numbers. However, that doesn't seem to simplify
anything. Now instead of assuming the list contains all reals,
we must make no assumptions. But that means we must
consider incomplete lists and complete lists, because the
construction that occurs in step 2 must be shown to be valid
in all cases. In the case of lists that contain all reals, which
we must consider unless we are assuming no such list exists
(What good is the proof in that case?), then we immediately
recognize that the real number described in step #2 as
ill-defined and meaningless. So if we consider ANY list we
end up with the same problem. In addition we have another
case to consider, although I think nobody would object if we
neglected to consider that case here.
Arturo Magidin
As I understand it:
The "proof by contradiction" blueprint is basically that
[A -> (B and ~B)] -> ~A.
(or, in some forms, that (P -> ~P) -> ~P )
To be precise, in proof by contradiction, from an assumption A you
derive a contradiction, and you deduce ~(A). This is accepted by
constructivist. You can assume something exists, for example, deduce a
contradiction, and conclude that such a thing does not exist.
In the case where A is itself a negation, A = ~B, the constructivist
accepts a that from if from A you deduce a contradiction ("deduce"
here is restricted to methods allowed by constructivists, of course),
then you can conclude ~A. For most mathematicians, this is equivalent
to B, since ~A = ~~B, and excluded middle will give you B. For
constructivist this is not the case.
So: if you assume that there does not exist x such that P(x), and
deduce a contradiction, the constructivist will accept that "it is
absurd that there does not exist x such that P(x)". He will not,
however, accept that there ->exists<- an x such that P(x). i.e., he
will accept
~(~(Ex P(x))),
but he will not accept
Ex P(x).
>I understood that constructivists did not much care for the law of the
>excluded middle.
Arturo Magidin
<the_...@yahoo.com> wrote:
>Most of the responses point out that a simpler proof or a proof
>that avoids the problem of the self-reference is to consider ANY
>list of real numbers. However, that doesn't seem to simplify
>anything. Now instead of assuming the list contains all reals,
>we must make no assumptions. But that means we must
>consider incomplete lists and complete lists, because the
>construction that occurs in step 2 must be shown to be valid
>in all cases.
There is no need to consider any kind of list separately. Step 2 can
be shown to be valid for ANY function f:N->R, regardless of whether it
is one-to-one or not, regardless of whether we assume it to be
surjective or not. All we need to assume is that it is a ->function<-.
> In the case of lists that contain all reals, which
>we must consider unless we are assuming no such list exists
>(What good is the proof in that case?)
That you ask the question suggests you do not understand the argument
at all.
If one proves that for EVERY function f:N->R, there exists x in R (which
depends on f) such that x is not in f(N), then this proves that EVERY
list of real numbers is incomplete. This proves that NO list can be
complete, and that's what you want the proof for in the first place.
> then we immediately
>recognize that the real number described in step #2 as
>ill-defined and meaningless.
If you say so. Sure. Wanna buy a bridge?
Poker Joker
news:efhkpt$3136$1...@agate.berkeley.edu...
>> In the case of lists that contain all reals, which
>>we must consider unless we are assuming no such list exists
>>(What good is the proof in that case?)
>
> That you ask the question suggests you do not understand the argument
> at all.
It was rhetorical. Sometimes people ask questions that they don't
really expect to be answered. They do so when they are truing to make
a point. Wikipedia has a whole section on it if you would like to learn
more.
Poker Joker
> In article <1159410937.0...@h48g2000cwc.googlegroups.com>,
> <the_...@yahoo.com> wrote:
>>Cantor's proof is one of the most popular topics on this NG. It
>>seems that people are confused or uncomfortable with it, so
>>I've tried to summarize it to the simplest terms:
>>
>>1. Assume there is a list containing all the reals.
>>2. Show that a real can be defined/constructed from that list.
>>3. Show why the real from step 2 is not on the list.
>>4. Conclude that the premise is wrong because of the contradiction.
>
> This is hardly the simplest terms. Much simpler is to do a ->direct<-
> proof instead of a proof by contradiction.
>
> 1. Take ANY list of real numbers.
> 2. Show that a real can be defined/constructed from that list.
> 3. Show that the real from step 2 is not on the list.
> 4. Conclude that no list can contain all reals.
>
How can it be simpler if the list can be ANY list instead of a
particular one. ANY list opens up more possiblities than
a single list. Also, if its true for ANY list, then it must be
true for a specific list. So if considering a single specific list
shows a flaw, then looking at ANY (ALL of them) list doesn't
help.
Poker Joker
"Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
news:451b5130$0$28952$afc3...@news.optusnet.com.au...
> You produce ANY list of all Reals, I can show you a missing real.
> Therefore I can do it for ALL lists, and hence there is no complete list
> of Reals.
If he shows you a list of all reals you can't. I thought that would be
obvious
to even the casual observer.
cbr...@cbrownsystems.com
Virgil wrote:
> In article <1159417542.4...@i3g2000cwc.googlegroups.com>,
> cbr...@cbrownsystems.com wrote:
>
> > the_...@yahoo.com wrote:
> > > Cantor's proof is one of the most popular topics on this NG. It
> > > seems that people are confused or uncomfortable with it, so
> > > I've tried to summarize it to the simplest terms:
> > >
> > > 1. Assume there is a list containing all the reals.
> > > 2. Show that a real can be defined/constructed from that list.
> > > 3. Show why the real from step 2 is not on the list.
> > > 4. Conclude that the premise is wrong because of the contradiction.
>
> That is a proof by contradiction, which many constructionists object to.
>
I assume you are replying to the OP here, not me.
Cheers - Chas
Ross A. Finlayson
That illustrates a differentiation between me and Dave. Where Dave
considers himself an anti-Cantorian, and negates in denial for
application of otherwise suitable incompatible results the powerset
result, I'm a post-Cantorian, and obviate the powerset result. Is that
fair, Dave?
Dave, Dave, I suggest to staunch Cantorians that they print a list of
the naturals, just, not here, because I prefer to write it as N.
If you use a (the) universe in your set theory, it's not ZF, and the
powerset result doesn't universally hold. Successor is power type.
Universe: infinite; infinite sets: equivalent.
Ross
Poker Joker
> If one proves that for EVERY function f:N->R, there exists x in R (which
> depends on f) such that x is not in f(N), then this proves that EVERY
> list of real numbers is incomplete. This proves that NO list can be
> complete, and that's what you want the proof for in the first place.
Why the switch to functional notation? I think you are trying to make it
more complicated than it is.
The OP already pointed out the problem with your argument. The point
is that you ->CAN'T<- prove that for ->EVERY<- function unless you
assume that ->NONE<- of them have R as their image. And since you
must assume that ->SOME MIGHT<- have R as their image, there
exists no such real number because the definition of that real is
self-referential in that case.
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
> news:451b5130$0$28952$afc3...@news.optusnet.com.au...
>
> > You produce ANY list of all Reals, I can show you a missing real.
> > Therefore I can do it for ALL lists, and hence there is no complete list
> > of Reals.
>
> If he shows you a list of all reals you can't.
The point is that if there is no such list he cannot show you one, and
the proof show that there is no such list.
> I thought that would be
> obvious
> to even the casual observer.
My very words!
Virgil
cbr...@cbrownsystems.com wrote:
Right! Sorry, should have sniped out your ID to make it clear I as only
re[plying to the_...@yahoo.com
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
> "Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
> news:efhkpt$3136$1...@agate.berkeley.edu...
>
> > If one proves that for EVERY function f:N->R, there exists x in R (which
> > depends on f) such that x is not in f(N), then this proves that EVERY
> > list of real numbers is incomplete. This proves that NO list can be
> > complete, and that's what you want the proof for in the first place.
>
> Why the switch to functional notation? I think you are trying to make it
> more complicated than it is.
>
> The OP already pointed out the problem with your argument. The point
> is that you ->CAN'T<- prove that for ->EVERY<- function unless you
> assume that ->NONE<- of them have R as their image.
Maybe you can't but Arturo can, and Cantor could, as that's the way it
was originally done.
> And since you
> must assume that ->SOME MIGHT<- have R as their image
One need not assume any such thing.
One proves directly that for every list there exists a number not listed
in that list.
Randy Poe
Poker Joker wrote:
> "Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
> news:efhkpt$3136$1...@agate.berkeley.edu...
>
> > If one proves that for EVERY function f:N->R, there exists x in R (which
> > depends on f) such that x is not in f(N), then this proves that EVERY
> > list of real numbers is incomplete. This proves that NO list can be
> > complete, and that's what you want the proof for in the first place.
>
> Why the switch to functional notation?
To make the statement more compact.
> I think you are trying to make it
> more complicated than it is.
It was one sentence. How complicated is that?
> The OP already pointed out the problem with your argument. The point
> is that you ->CAN'T<- prove that for ->EVERY<- function unless you
> assume that ->NONE<- of them have R as their image.
That's incorrect. You don't have to assume none map onto R in order to
prove none map onto R.
The direct argument starts this way: Let f be any such function, from
naturals to reals.
Now, are you saying that somehow that misses some possible functions
from naturals to reals? How so?
Given only that assumption about f (which includes ALL POSSIBLE
SUCH FUNCTIONS) it follows that f misses some reals. It follows
just from assuming f maps naturals to reals.
What function from naturals to reals is missed by that argument?
- Randy
cbr...@cbrownsystems.com
Apply your logic to "oddven".
A natural number is even if it is of the form 2*n, and odd if it is not
of this form. A number is oddven if it is both even and odd.
1) Assume that n is oddven.
2) If it is oddven, then it is even; so it is of the form 2*n.
3) But if it is oddven, then it is odd; and therefore it is not of the
form 2*n.
4) Contradiction; therefore no such n exists.
Simpler is the following:
1) Let n be any even number.
2) n is then of the form 2*n.
3) It follows that n cannot be odd.
4) Therefore, there is no such thing as an oddven number.
In the case of the original argument, substitute "is a list of real
numbers" for "even", "is a complete mapping" for "odd", and "is a
complete list of the reals" for "oddven".
Cheers - Chas
Poker Joker
"Virgil" <vir...@comcast.net> wrote in message
news:virgil-1201BC....@comcast.dca.giganews.com...
> In article <C2ZSg.1209$3E2...@tornado.rdc-kc.rr.com>,
> "Poker Joker" <Po...@wi.rr.com> wrote:
>
>> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
>> news:451b5130$0$28952$afc3...@news.optusnet.com.au...
>>
>> > You produce ANY list of all Reals, I can show you a missing real.
>> > Therefore I can do it for ALL lists, and hence there is no complete
>> > list
>> > of Reals.
>>
>> If he shows you a list of all reals you can't.
>
> The point is that if there is no such list he cannot show you one, and
> the proof show that there is no such list.
The point is that if there is such a list the proof that there isn't one
isn't valid, but even more, under the assumption that there *MAY*
be such a list, the proof is not valid because it would entail a
self-referential definition.
Whether the proof is by-contradiction or not is immaterial. Either
way the real number from step #2 is defined in terms of itself under
the assumption that the list *MIGHT* contain all the reals.
Poker Joker
<cbr...@cbrownsystems.com> wrote in message
news:1159495306....@e3g2000cwe.googlegroups.com...
> Apply your logic to "oddven".
>
> A natural number is even if it is of the form 2*n, and odd if it is not
> of this form. A number is oddven if it is both even and odd.
>
> 1) Assume that n is oddven.
> 2) If it is oddven, then it is even; so it is of the form 2*n.
> 3) But if it is oddven, then it is odd; and therefore it is not of the
> form 2*n.
> 4) Contradiction; therefore no such n exists.
First, the OP isn't saying anything is wrong with "proofs
by contradiction". He's saying that there is a meaningless
definition like "This statement is false."
There is no meaningless definition in your step 2.
Defining a real number in terms of all real numbers in a
set is self referential if the set contains all the real
numbers. The same is true for lists. So under the
assumption that the list *MIGHT* contain all the real
numbers, we can't say that the definition is not
self-referential. Therefore it is meaningless.
William Hughes
Nope. The steps are as follows.
Consider a list of real numbers L. (At this point we do not know
if L contains all real numbers or not).
In your terminology, L might contain all real numbers.
Define a procedure D, that takes a list M and produces
a real number r=D(M). This procedure D will work on any
list M
Apply the procedure D to the list L to get a real
number r=D(L). Note that D(L) exists because we can
apply D to any list. The fact that L *might* contain all
the real numbers is not a problem because D was
defined in such a way that it can be applied to any list.
We now have a list L and a real number r. We
argue that by the way r was constructed r cannot
be an element of L. At this point we conclude
that L does not contain all the real numbers.
There is nothing self referential here.
- William Hughes
Poker Joker
"Randy Poe" <poespa...@yahoo.com> wrote in message
news:1159494111....@i3g2000cwc.googlegroups.com...
> That's incorrect. You don't have to assume none map onto R in order to
> prove none map onto R.
>
> The direct argument starts this way: Let f be any such function, from
> naturals to reals.
Certainly we should assume that f *MIGHT* have R as its image, right?
> Now, are you saying that somehow that misses some possible functions
> from naturals to reals? How so?
No, but we haven't proven that the image of f can't be R in step #1, right?
So step #2 isn't valid, right? It's only *VALID* if R isn't the image of f.
But without step #2 being valid under *ALL* conditions, the proof
as a whole breaks down.
> Given only that assumption about f (which includes ALL POSSIBLE
> SUCH FUNCTIONS) it follows that f misses some reals. It follows
> just from assuming f maps naturals to reals.
Under the most general assumption, we can't count out that
R is f's image, so defining a real in terms of the image of
f *MIGHT* be self-referential, and it certainly is if the image
of f is R. So, in general, the best we can say is that the
real from step #2 might be self-referential and hasn't been
shown not to be so without assuming the conclusion is true.
I don't usually accept proofs that *REQUIRE* you to assume
the conclusion is true for the proof to be valid. At best, the proof
might show the conclusion to be independent of the premises.
> What function from naturals to reals is missed by that argument?
I'm not missing any, because I'm considering functions with R as
their image. It seems that you are the one not considering
those functions and their implication for step #2.
Poker Joker
"William Hughes" <wpih...@hotmail.com> wrote in message
news:1159499239.8...@c28g2000cwb.googlegroups.com...
> Define a procedure D, that takes a list M and produces
> a real number r=D(M). This procedure D will work on any
> list M
If M contains all reals, it contains r, so:
r = D( M : r in M)
See the self reference?
William Hughes
No
I suppose what you mean is
r = D(M, with M being a list that contains r)
but we never have this. What we have is
r = D(M, with M being a list that may contain r)
later we learn that M did not contain r (no contradiction we
only said may).
- William Hughes
Ross A. Finlayson
Poker Joker:
Bravo!
You're damn right.
Dang, I wanted to be 39. I must have seen ten thousand usenet
messages.
Ross
cbr...@cbrownsystems.com
> "William Hughes" <wpih...@hotmail.com> wrote in message
> news:1159499239.8...@c28g2000cwb.googlegroups.com...
>
> > Define a procedure D, that takes a list M and produces
> > a real number r=D(M). This procedure D will work on any
> > list M
>
Likewise, the definition given implies that M is a complete list of
reals if and only if M is complete, and M is a list of reals.
If M is a list of reals, we can construct a real which is not in M;
i.e, M is not complete.
Therefore, the assertion "M is a complete list of reals" is only true
if the assertion "M is complete, and M is not complete" is true.
(A and ~A) = false.
Therefore, the assertion "M is a complete list of reals" is logically
false for any M. There is no need to assume that an object such as a
complete list of all reals exists in order to prove that such an object
does not exist.
Cheers - Chas
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
> "Virgil" <vir...@comcast.net> wrote in message
> news:virgil-1201BC....@comcast.dca.giganews.com...
> > In article <C2ZSg.1209$3E2...@tornado.rdc-kc.rr.com>,
> > "Poker Joker" <Po...@wi.rr.com> wrote:
> >
> >> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
> >> news:451b5130$0$28952$afc3...@news.optusnet.com.au...
> >>
> >> > You produce ANY list of all Reals, I can show you a missing real.
> >> > Therefore I can do it for ALL lists, and hence there is no complete
> >> > list
> >> > of Reals.
> >>
> >> If he shows you a list of all reals you can't.
> >
> > The point is that if there is no such list he cannot show you one, and
> > the proof show that there is no such list.
>
> The point is that if there is such a list the proof that there isn't one
> isn't valid, but even more, under the assumption that there *MAY*
> be such a list, the proof is not valid because it would entail a
> self-referential definition.
>
> Whether the proof is by-contradiction or not is immaterial. Either
> way the real number from step #2 is defined in terms of itself under
> the assumption that the list *MIGHT* contain all the reals.
Maybe in Jokerland, but not in any real world.
If, as has been proved, EVERY list leaves out at least one real, then NO
list can possible list all of them.
Your objections have no logic behind them, and in, fact, contradict the
status quo.
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
One need not make that assumption. One need only refrain from assuming
it to be impossible.
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1159494111....@i3g2000cwc.googlegroups.com...
>
> > That's incorrect. You don't have to assume none map onto R in order to
> > prove none map onto R.
> >
> > The direct argument starts this way: Let f be any such function, from
> > naturals to reals.
>
> Certainly we should assume that f *MIGHT* have R as its image, right?
Why make any assumption at all?
Virgil
Eric Schmidt
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1159494111....@i3g2000cwc.googlegroups.com...
>
>
>>That's incorrect. You don't have to assume none map onto R in order to
>>prove none map onto R.
>>
>>The direct argument starts this way: Let f be any such function, from
>>naturals to reals.
>
>
> Certainly we should assume that f *MIGHT* have R as its image, right?
Why? What does it even mean to assume something might be true? I could
"assume" that Goldbach's Conjecture might be true, but that's really not
assuming anything substantial.
>>Now, are you saying that somehow that misses some possible functions
>>from naturals to reals? How so?
>
>
> No, but we haven't proven that the image of f can't be R in step #1, right?
> So step #2 isn't valid, right? It's only *VALID* if R isn't the image of f.
> But without step #2 being valid under *ALL* conditions, the proof
> as a whole breaks down.
Nothing in step 2 (or anything else) assumes or requires that R isn't
the image of f.
>>Given only that assumption about f (which includes ALL POSSIBLE
>>SUCH FUNCTIONS) it follows that f misses some reals. It follows
>>just from assuming f maps naturals to reals.
>
>
> Under the most general assumption, we can't count out that
> R is f's image, so defining a real in terms of the image of
> f *MIGHT* be self-referential, and it certainly is if the image
> of f is R. So, in general, the best we can say is that the
> real from step #2 might be self-referential and hasn't been
> shown not to be so without assuming the conclusion is true.
> I don't usually accept proofs that *REQUIRE* you to assume
> the conclusion is true for the proof to be valid. At best, the proof
> might show the conclusion to be independent of the premises.
Defining a real from a set of reals is not self-referential, even if the
set contains all reals. The fact that the real may be in the set does
not make the definition self-referential. For example, consider a
function g whose domain is the set of all subsets of R. For S a subset
of R, g(S) is the supremum of S if S is non-empty and bounded above.
Otherwise, g(S) = 0. Do you assert that this definition is
self-referential if S = R? If S = {x in R : x <= 0}?
>>What function from naturals to reals is missed by that argument?
>
>
> I'm not missing any, because I'm considering functions with R as
> their image. It seems that you are the one not considering
> those functions and their implication for step #2.
You have not given any coherent explanation of why such functions are a
special case.
--
Eric Schmidt
--
Posted via a free Usenet account from http://www.teranews.com
Tonico
This is huge nonsense. The functional notation is rather handy when
trying to talk mathematics: mimicry and poetry not necessarily help
here.
The argument seems extremely simple to me...so simple that some seem to
feel it MUST be wrong, for some reason (perhaps the reason is that if
something ain't complex then it can't be maths...): take ANY list of
real numbers (or take ANY injective mapping
f:N --> R). Then it can be shown with the genial and simple diagonal
argument that Cantor came up, that some real is missing there (that
there's some x in R s.t. x doesn't belong to
f(N)).
Tell it this way, tell it that way: all is correct mathematicalwise
(?!) unless proved mathematicalwise (?!) that it isn't....someone wanna
buy a bridge AND the river under it?
Tonio
Han de Bruijn
> What a fallacious imbecile you are, Muckenschleim. Why don't you go
> drown your sorry feeble mind in beer at the Oktoberfest? Maybe they
> will beat the crap out of you for good there.
Hey, hey ! Would you please mind that this is a mathematics newsgroup
and that you are bringing _mathematics_ into disrepute with this kind
of uncivilized behaviour ? I've always thought that mathematics is a
form of civilization. But it seems that mainstream mathematics is NOT.
Han de Bruijn
Randy Poe
Poker Joker wrote:
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1159494111....@i3g2000cwc.googlegroups.com...
>
> > That's incorrect. You don't have to assume none map onto R in order to
> > prove none map onto R.
> >
> > The direct argument starts this way: Let f be any such function, from
> > naturals to reals.
>
> Certainly we should assume that f *MIGHT* have R as its image, right?
Uh, no. We just assume that f is a function that takes naturals
and produces reals.
For instance, let f(x) = sqrt(x).
Do I have to assume that f(x) "might have R as its image" in order
to talk about f(x) = sqrt(x)? Can't I just talk about f(x) = sqrt(x)
and examine what properties is has rather than speculating about
what it might have?
- Randy
Dik T. Winter
> In article <j%WSg.22037$SY6....@reader1.news.jippii.net>,
> Aatu Koskensilta <aatu.kos...@xortec.fi> wrote:
...
> > > That is a proof by contradiction, which many constructionists object to.
> >
> > constructivistic proof of ~A is a proof of contradiction from A. What
> > one can't do constructively is to prove A by proving that ~A leads to a
> > contradiction.
>
> In my understanding, a "proof by contradiction" is essentially a proof
> of A by proving ~A false, which requires accepting the law of the
> excluded middle.
This version they indeed do object to. But a "proof by contradiction"
can also be a proof of ~A by proving A false, which they do not object
to. In constructivist sense, when you proof that ~A leads to a
contradiction, you have proven ~~A, but not necessarily A.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Dik T. Winter
>
> "Randy Poe" <poespa...@yahoo.com> wrote in message
> news:1159494111....@i3g2000cwc.googlegroups.com...
>
> > That's incorrect. You don't have to assume none map onto R in order to
> > prove none map onto R.
> >
> > The direct argument starts this way: Let f be any such function, from
> > naturals to reals.
>
> Certainly we should assume that f *MIGHT* have R as its image, right?
You may assume that, but that assumption is not needed.
> > Now, are you saying that somehow that misses some possible functions
> > from naturals to reals? How so?
>
> No, but we haven't proven that the image of f can't be R in step #1, right?
> So step #2 isn't valid, right?
Remember:
> 1. Assume there is a list containing all the reals.
> 2. Show that a real can be defined/constructed from that list.
> 3. Show why the real from step 2 is not on the list.
> 4. Conclude that the premise is wrong because of the contradiction.
Why is step 2 invalid?
> Under the most general assumption, we can't count out that
> R is f's image, so defining a real in terms of the image of
> f *MIGHT* be self-referential, and it certainly is if the image
> of f is R.
What is the problem here?
Arturo Magidin
>> In article <1159410937.0...@h48g2000cwc.googlegroups.com>,
>> <the_...@yahoo.com> wrote:
>>>Cantor's proof is one of the most popular topics on this NG. It
>>>seems that people are confused or uncomfortable with it, so
>>>I've tried to summarize it to the simplest terms:
>>>
>>>2. Show that a real can be defined/constructed from that list.
>>>3. Show why the real from step 2 is not on the list.
>>>4. Conclude that the premise is wrong because of the contradiction.
>>
>> proof instead of a proof by contradiction.
>>
>> 1. Take ANY list of real numbers.
>> 2. Show that a real can be defined/constructed from that list.
>> 4. Conclude that no list can contain all reals.
>>
>
>How can it be simpler if the list can be ANY list instead of a
>particular one.
Because a direct proof is simpler than a proof by contradiction.
> ANY list opens up more possiblities than
>a single list.
Any list does not require you to assume that there is a "single list"
which some some particular property.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Arturo Magidin
Poker Joker <Po...@wi.rr.com> wrote:
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
>news:efhkpt$3136$1...@agate.berkeley.edu...
>
>> If one proves that for EVERY function f:N->R, there exists x in R (which
>> depends on f) such that x is not in f(N), then this proves that EVERY
>> list of real numbers is incomplete. This proves that NO list can be
>> complete, and that's what you want the proof for in the first place.
>
>Why the switch to functional notation?
To make clear and explicit was has up to now been implicit: that a
list is a function from the natural numbers to the reals.
> I think you are trying to make it
>more complicated than it is.
You are free to think what you want. Reality notwithstanding.
>The OP already pointed out the problem with your argument.
No, the original poster pointed out what he thinks is a problem with
the proof.
> The point
>is that you ->CAN'T<- prove that for ->EVERY<- function unless you
>assume that ->NONE<- of them have R as their image.
Using capital letters does not make the objection true.
Your inability to prove it does not affect my ability to prove that
for every function function f:N->R, assuming only that it is a
function from N to R, and no other properties, there exists x such
that x is not in f(N). Since the proof assumes nothing at all about f
other than the fact that it is a function, the proof shows that I can
prove it for every function, and that I do not assume anything like
what you assert is being assumed.
>And since you
>must assume that ->SOME MIGHT<- have R as their image,
I assume nothing about the range of f, other than that it is contained
in R. Whether or not it has R as the image is irrelevant.
georgie
Randy Poe wrote:
>
> Uh, no. We just assume that f is a function that takes naturals
> and produces reals.
>
> For instance, let f(x) = sqrt(x).
>
> Do I have to assume that f(x) "might have R as its image" in order
> to talk about f(x) = sqrt(x)? Can't I just talk about f(x) = sqrt(x)
> and examine what properties is has rather than speculating about
> what it might have?
No, but f(x) = sqrt(x) isn't true in general. We don't take it as
being
valid for all x.
georgie
Sorry, I mean x = sqrt(x).
Jesse F. Hughes
I have never seen such stupid Jingoism expressed in a mathematics
journal. Why would you think that guenther's post is representative
of mainstream mathematics?
Mathematics, like every field, has enthusiasts and practitioners that
are quite stupid. No field is immune from ignorance. That does not
mean that standard mathematical practice is characterized by
ignorance.
--
"My personal opinion is, it was a shameful act for someone to disclose
this very important program in a time of war. The fact that we're
discussing this program is helping the enemy."
-- George W. Bush: Questioning Illegal Wiretaps Just Helps The Enemy!
Jesse F. Hughes
> Each sequence of real numbers omits at least one real
> number. (Sequence being a 1-1 and onto function from the positive
> integers to the real numbers.)
Er, are you sure about that definition of sequence? Seems to me that
a sequence is any function from the positive integers to the reals.
With your definition, the conclusion would be: every onto function
N->R is not onto. This is true, of course, but is more likely to
confuse the "Anti-Cantorians" rather than enlighten them.
--
Jesse F. Hughes
"I guess it's a passable day to die."
-- Lt. Dwarf, /Star Wreck:In the Pirkinning"
Jesse F. Hughes
> The argument of the "anti-Cantorians" is that "real" mathematics is
> computationally testable.
Bullshit. That's *your* argument. Perhaps some other so-called
"anti-Cantorians" have agreed with you, but that is not the motivation
of every anti-Cantorian.
--
"So, at this time, I'd like to assure you that I am not interested in
making sure mathematicians worldwide get fired."--JSH Apr 28, 2003
"I'll have prosecutors knocking on your doors. I have no problem with
any number of mathematicians spending time in jail."--JSH Jun 10, 2003
Han de Bruijn
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
>>guenther vonKnakspot wrote:
>>
>>>What a fallacious imbecile you are, Muckenschleim. Why don't you go
>>>drown your sorry feeble mind in beer at the Oktoberfest? Maybe they
>>>will beat the crap out of you for good there.
>>
>>Hey, hey ! Would you please mind that this is a mathematics
>>newsgroup and that you are bringing _mathematics_ into disrepute
>>with this kind of uncivilized behaviour ? I've always thought that
>>mathematics is a form of civilization. But it seems that mainstream
>>mathematics is NOT.
>
> I have never seen such stupid Jingoism expressed in a mathematics
> journal. Why would you think that guenther's post is representative
> of mainstream mathematics?
>
> Mathematics, like every field, has enthusiasts and practitioners that
> are quite stupid. No field is immune from ignorance. That does not
> mean that standard mathematical practice is characterized by
> ignorance.
I don't remember I was saying something about "ignorance". I thought
I was saying something about rudeness / uncivilized behaviour.
Han de Bruijn
Albrecht
the_...@yahoo.com schrieb:
> Cantor's proof is one of the most popular topics on this NG. It
> seems that people are confused or uncomfortable with it, so
> I've tried to summarize it to the simplest terms:
>
> 1. Assume there is a list containing all the reals.
> 2. Show that a real can be defined/constructed from that list.
> 3. Show why the real from step 2 is not on the list.
> 4. Conclude that the premise is wrong because of the contradiction.
>
> The steps are simple except for a possible debate about defined /
> constructed. I don't think anyone believes the proof is invalid
> because of that debate however.
>
> There seems to be another area that seems to be a problem
> though. The problem is that step #2 doesn't seem valid. If we
> assume the list contains all the real numbers, then defining or
> constructing a real number in terms of that list would be
> self-referential. The number from step #2, that is normally defined
> digit-by-digit along the diagonal, must have its digits (or at least
> one of them) defined as not equal to itself, if we are to assume the
> list contains all the real numbers. Certainly the conclusion in that
> case is that the premise is wrong or that the construction is not
> valid, but the conclusion can't be simply that the premise is wrong.
>
> This same problem appears in the "power-set" theorem, where we
> have a definition of a set, S, which is a subset of N, defined in
> terms of the image of a function, f, whose image is assumed to be
> the power-set of N. If the image of f is assumed to be the power-set
> of N and S is defined in terms of f, then S must necessarily be
> defined in terms of itself. Again, if we assume that the image of
> f is P(N), then defining S as a set whose elements are defined
> to be elements not in the image of f is a self-referential definition
> of S because S is also a subset of N, making it meaningless.
> Certainly a meaningless definition can't be used to prove a
> contradiction.
>
> I'm guessing that the "discussions" that occur stem from the fact
> that mathematicians disagree that the seemingly self-referential
> definitions are a problem but it's not intuitively obvious why that is
> so, therefore many people feel the need to try to refute the proofs.
> The problem is that it really isn't clear why mathematicians seem to
> accept the self-referential definitions.
Cantor's argument shows just one only thing:
there is no list of all reals - as there is no list of all naturals -
as there isn't anything which is infinite and completed.
To ask about a real which should be constructed from an infinite list
is as to ask after the sum of all natural numbers.
There is no sum of all natural numbers as there is no real which is
constructed from all elements of an infinite list.
Best regards
Albrecht S. Storz
guenther vonKnakspot
Jesse F. Hughes wrote:
> Han de Bruijn <Han.de...@DTO.TUDelft.NL> writes:
>
> > guenther vonKnakspot wrote:
> >
> >> What a fallacious imbecile you are, Muckenschleim. Why don't you go
> >> drown your sorry feeble mind in beer at the Oktoberfest? Maybe they
> >> will beat the crap out of you for good there.
> >
> > Hey, hey ! Would you please mind that this is a mathematics
> > newsgroup and that you are bringing _mathematics_ into disrepute
> > with this kind of uncivilized behaviour ? I've always thought that
> > mathematics is a form of civilization. But it seems that mainstream
> > mathematics is NOT.
>
> I have never seen such stupid Jingoism expressed in a mathematics
> journal. Why would you think that guenther's post is representative
> of mainstream mathematics?
manner my reply to Muckenschleim is jingoistic, or how it is
representative of any kind of mathematics at all.
> Mathematics, like every field, has enthusiasts and practitioners that
> are quite stupid. No field is immune from ignorance. That does not
> mean that standard mathematical practice is characterized by
> ignorance.
Hey, how come you can call me stupid but I can't call Muckenschleim
fallacious imbecile which he demonstrably is?
Regards
William Hughes
A set does not have to be infinite to be countable.
Cantor's proof works just fine for finite sets. So
Cantor's proof can be used to show that the set
of real numbers is not finite. [If this is not cracking a walnut
with a sledgehammer, I don't know what is!]
If you do not allow "completed infinite sets", then the meat
of Cantor's proof (the cardinality of the reals is greater than
that of the integers) cannot be done
So, everyone who finds your "contradictions" to be convincing
will find Cantor rather trivial.
However, as far as I can see, 'everyone who finds your
"contradictions" to be convincing', is a set with at most
one element.
-William Hughes
guenther vonKnakspot
payed by taxpayers. He inflicts his fallacious crap upon real students
in full awareness of its incorrectness. There is absolutely nothing
civilized about that. There is nothing about Muckenschleim to elicit
civilized behaviour when dealing with him.
Regards.
guenther vonKnakspot
Hey moron, give us your definition of "list" under which there is no
list of all naturals.
Randy Poe
What do you mean "valid"? That's my definition of f(x).
For any x, what I mean by f(x) is the value of sqrt(x).
x in this context is a natural number, so sqrt(x) is a
positive real for all x.
- Randy
Dave L. Renfro
>> Each sequence of real numbers omits at least one real
>> number. (Sequence being a 1-1 and onto function from the
>> positive integers to the real numbers.)
Jesse F. Hughes wrote:
> Er, are you sure about that definition of sequence? Seems to
> me that sequence is any function from the positive integers
> to the reals.
>
> With your definition, the conclusion would be: every onto
> function N->R is not onto. This is true, of course, but is
> more likely to confuse the "Anti-Cantorians" rather than
> enlighten them.
Ooops, you're correct ... I messed up. I knew I should have
stayed out of one of these kinds of threads. I was curious
how mueckenh would respond to my bringing up other situations
(in "real life") where one argues that something can't happen
by showing that something contradictory (or at least, something
not desirable) would arise if we assumed it did happen. He (she?)
seemed to have ignored my examples, though.
And what's with this other thing I saw in the thread, where
someone argued that the diagonal argument is not complete
because the assumption that the list of real numbers containing
all the real numbers wasn't allowed for? That's just bizarre.
If this person is _really_ concerned about that issue (rather
than trolling, as I strongly suspect), why isn't he raising
the same issue for _every_ argument? For example, in proving
that 4 + 3 = 7 (using three applications of the successor
function), he should complain that we didn't consider the
case where 4 + 3 isn't 7, and when we consider this possibility,
the proof fails. More generally, he should find fault with
every proof of statements of the form "If P, then Q", including
the arguments that he is using to support his position.
Now that I think about it, don't a lot of these anit-Cantor
arguments take the same form as what they're criticizing?
They argue that something about diagonalizing is incorrect,
because if it were correct, then [insert their attempt to
obtain a contradiction].
Dave L. Renfro
Ross A. Finlayson
Hi, Dave,
I think it's more interesting to try and generate lists, or mappings,
in certain ways and then observe the construction of the anti-diagonal.
I don't care for calling the anti-diagonal argument the diagonal
argument.
There are basically anti-diagonal and powerset results to consider, and
nested intervals, those comprise the core of what are called Cantorian
results.
With the anti-diagonal result, it's interesting to see how different
bases of the numbers can provide different restrictions on the rules in
selection of anti-diagonals. For example, you'll notice that the
anti-diagonal argument is generally presented in base four or higher,
where four resolves to two and etc., and not in some infinite base.
With f(x) = x+1 and the powerset result, over the naturals as domain
the missing element is the empty set. If your naturals don't have zero
being the empty set, then there are some considerations possible about
vacuity, the empty set, and how the empty set is a subset of any other
set, besides itself, because it contains no elements not in the set,
not because it contains elements from the set.
There is no universe in ZF. The universe as a set in a set theory is
its own powerset, violating the power set proof which does not
necessarily apply to irregular sets. Where all infinite sets are
irregular, hell, where all sets are irregular, the finite combinatorics
results still are perfect.
Then again, I'm post-Goedelian too.
Ross
Herman Jurjus
[snip]
> This version they indeed do object to. But a "proof by contradiction"
> can also be a proof of ~A by proving A false, which they do not object
> to.
With interest i observe that you people here on sci.math talk about
constructivists in terms of 'us' and 'them'.
;-)
--
Cheers,
Herman Jurjus
A N Niel
Jurjus <h.ju...@hetnet.nl> wrote:
>
> With interest i observe that you people here on sci.math talk about
> constructivists in terms of 'us' and 'them'.
>
And I observe that Herman talks about "you people here" as though he
were not here as well.
Poker Joker
>
> georgie wrote:
>> Randy Poe wrote:
>> >
>> > Uh, no. We just assume that f is a function that takes naturals
>> > and produces reals.
>> >
>> > For instance, let f(x) = sqrt(x).
>> >
>> > Do I have to assume that f(x) "might have R as its image" in order
>> > to talk about f(x) = sqrt(x)? Can't I just talk about f(x) = sqrt(x)
>> > and examine what properties is has rather than speculating about
>> > what it might have?
>>
>> No, but f(x) = sqrt(x) isn't true in general. We don't take it as
>> being valid for all x.
>
> What do you mean "valid"? That's my definition of f(x).
>
> For any x, what I mean by f(x) is the value of sqrt(x).
Thats not the same as defining x in terms of the set of all reals.
Poker Joker
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes" by Bill Watterson)
> ======================================================================
We all noticed you neglected this logic:
if its true for ANY list, then it must be
true for a specific list. So if considering a single specific list
shows a flaw, then looking at ANY (ALL of them) list doesn't
help.
Poker Joker
"Dik T. Winter" <Dik.W...@cwi.nl> wrote in message
news:J6CsB...@cwi.nl...
> In article <070Tg.14143$8_5....@tornado.rdc-kc.rr.com> "Poker Joker"
> <Po...@wi.rr.com> writes:
> >
> > "Randy Poe" <poespa...@yahoo.com> wrote in message
> > news:1159494111....@i3g2000cwc.googlegroups.com...
> >
> > > That's incorrect. You don't have to assume none map onto R in order to
> > > prove none map onto R.
> > >
> > > The direct argument starts this way: Let f be any such function, from
> > > naturals to reals.
> >
> > Certainly we should assume that f *MIGHT* have R as its image, right?
>
> You may assume that, but that assumption is not needed.
Certainly not for ostriches.
> > > Now, are you saying that somehow that misses some possible functions
> > > from naturals to reals? How so?
> >
> > No, but we haven't proven that the image of f can't be R in step #1,
> > right?
> > So step #2 isn't valid, right?
>
> Remember:
> > 1. Assume there is a list containing all the reals.
> > 2. Show that a real can be defined/constructed from that list.
> > 3. Show why the real from step 2 is not on the list.
> > 4. Conclude that the premise is wrong because of the contradiction.
>
> Why is step 2 invalid?
Do you always accept steps that have questionable validity?
> > Under the most general assumption, we can't count out that
> > R is f's image, so defining a real in terms of the image of
> > f *MIGHT* be self-referential, and it certainly is if the image
> > of f is R.
>
> What is the problem here?
I assume you accept this proof that there are no complete lists
of reals:
Let r be a real number between 0 and 1. Let r_n denote the nth digit
in r's decimal expansion. Let r_n = 5 if r_n = 4, otherwise let r_n = 4.
r isn't on any list of reals. Therefore there isn't a complete list of
reals.
MoeBlee
> "Arturo Magidin" <mag...@math.berkeley.edu> wrote in message
> news:efgfhd$261u$1...@agate.berkeley.edu...
> > In article <1159410937.0...@h48g2000cwc.googlegroups.com>,
> > <the_...@yahoo.com> wrote:
> >>Cantor's proof is one of the most popular topics on this NG. It
> >>seems that people are confused or uncomfortable with it, so
> >>I've tried to summarize it to the simplest terms:
> >>
> >>2. Show that a real can be defined/constructed from that list.
> >>3. Show why the real from step 2 is not on the list.
> >>4. Conclude that the premise is wrong because of the contradiction.
> >
> > proof instead of a proof by contradiction.
> >
> > 1. Take ANY list of real numbers.
> > 4. Conclude that no list can contain all reals.
> >
>
> How can it be simpler if the list can be ANY list instead of a
> a single list.
By 'any' we mean an arbitrary one. The way we talk about an arbitrary
object is to choose a variable not free in any previous line in the
argument nor free in the conclusion we will eventually draw and then
use the rule of universal generalization to draw our eventual
conclusion. So if we want to prove something about an arbitrary
enumeration of denumerable sequences of digits, we say, "Suppose f is
an enumeration whose range is a subset of the set of denumerable
sequences of digits" (and, of course, we will presume NOTHING about f
other than that it is an enumeration whose range is a subset of the set
of denumerable sequences of digits.
> Also, if its true for ANY list, then it must be
> true for a specific list.
If the property holds for any list, then, if there exists a list, then
the property holds of any such list that exists.
> So if considering a single specific list
> shows a flaw, then looking at ANY (ALL of them) list doesn't
> help.
If there is a specific list that does not have the property, then we
will not be able to prove that the property holds of an arbitrary list.
But I don't know what specific list you think is "flawed". Nor do I see
what your point has to do with Arturo's point that we don't have to
adopt a reductio ad absurdum assumption, since we can just show for an
arbitrary list that it does not list every real number.
MoeBlee
Virgil
"Poker Joker" <Po...@wi.rr.com> wrote:
But the construction method, which applies equally well to all lists
without modification, if flawless for any list is equally flawless for
all lists.
And it is flawless for any list.
Poker Joker
"Virgil" <vir...@comcast.net> wrote in message
news:virgil-A8F506....@comcast.dca.giganews.com...
Define a real number r between 0 and 1. Denote
r_n to be the nth digit of r. The digits of r are defined
as follows:
r_n = 5 if r_n = 4, otherwise r_n = 4.
That cunstruction applies equally well to all lists
without modification. If it is flawless for any list it is
flawless for all lists.
And it is flawless for any list.
That's what you're saying. I see where you are
coming from and I won't bother answering such
"flawless" logic anymore, so don't bother.
MoeBlee
> "Peter Webb" <webbfamily...@optusnet.com.au> wrote in message
> news:451b5130$0$28952$afc3...@news.optusnet.com.au...
>
> > You produce ANY list of all Reals, I can show you a missing real.
> > Therefore I can do it for ALL lists, and hence there is no complete list
> > of Reals.
>
> If he shows you a list of all reals you can't. I thought that would be
> obvious
> to even the casual observer.
Of course. But so what? No one has shown an enumeration such that every
real number is in the range of the enumeration.
It's simple:
What we mean by a 'list of all reals' is a 1-1 function whose domain is
countable and whose range is the set of real numbers.
We show that for an arbitrary function f whose domain is countable,
there is some real number that is not in the range of f. Therefore, for
any arbitrary function f whose domain is countable, the range of f is
not the set of real numbers. So, since f is arbitrary, there is NO
function whose domain is countable and whose range is the set of real
numbers. A fortiori, there is no 1-1 function whose domain is
countable and whose range is the set of real numbers.
That's just plain logic, plain mathematical reasoning.
MoeBlee
Poker Joker
"MoeBlee" <jazz...@hotmail.com> wrote in message
news:1159574564....@k70g2000cwa.googlegroups.com...
By analogy, what you're saying is:
For ANY x
there is a procedure to find a y such that x/y = 1.
Because we are using the verbage "ANY", we don't
have to worry about special cases like when x = 0.
That's how mathematicians work?
Or are you just saying that you need not look at special
cases when we don't want to? Or is it that if a special
case is overlooked enough, then it no longer counts?
MoeBlee
>
> > If one proves that for EVERY function f:N->R, there exists x in R (which
> > depends on f) such that x is not in f(N), then this proves that EVERY
> > list of real numbers is incomplete. This proves that NO list can be
> > complete, and that's what you want the proof for in the first place.
>
> more complicated than it is.
No, he's not making it more complicated than it is. The functional
notation emphasizes that is FUNCTIONS we're talking about. A list is an
enumeration is a certain kind of function.
>
> you ->CAN'T<- prove that for ->EVERY<- function unless you
> assume that ->NONE<- of them have R as their image.
Wrong. We consider an arbitrary function from N into R and show that N
is not onto R. We do not assume that no function from N has R as the
range; rather we PROVE that no function from N has R as the range.
> And since you
> must assume that ->SOME MIGHT<- have R as their image, there
> exists no such real number because the definition of that real is
> self-referential in that case.
Whatever you mean by "self-referential", there is no step in the
argument that uses anything other than first order logic applied to the
axioms of set theory.
And we don't need to assume anything as to what "might" be the case. We
simply show that for an arbitrary function f whose domain is N, there
is some real number not in the range of f.
MoeBlee
MoeBlee
MoeBlee
> Whether the proof is by-contradiction or not is immaterial. Either
> way the real number from step #2 is defined in terms of itself under
> the assumption that the list *MIGHT* contain all the reals.
No, it is not. Either you are completely confused about a simple
mathematical argument, or you're just belching smoke that you know to
be smoke.
MoeBlee
Poker Joker
"Poker Joker" <Po...@wi.rr.com> wrote in message
news:WHiTg.25583$QT....@tornado.rdc-kc.rr.com...
>
> By analogy, what you're saying is:
>
> For ANY x
> there is a procedure to find a y such that x/y = 1.
>
> Because we are using the verbage "ANY", we don't
> have to worry about special cases like when x = 0.
> That's how mathematicians work?
>
> Or are you just saying that you need not look at special
> cases when we don't want to? Or is it that if a special
> case is overlooked enough, then it no longer counts?
Better yet:
For ANY real number x there is a procedure to find a real number y, such
that x/y = 1.
y isn't a real number when x = 0.
Therefore, 0 isn't a real number.
Poker Joker
Or you can't come up with anything better to say.
MoeBlee
> Defining a real number in terms of all real numbers in a
> set is self referential if the set contains all the real
> numbers. The same is true for lists. So under the
> assumption that the list *MIGHT* contain all the real
> self-referential. Therefore it is meaningless.
It's an arbitrary enumeration. We assume nothing about it other than
that it is an enumeration. We don't assume that it is an enumeration of
all real numbers and we don't assume that it is not an enumeration of
all real numbers. On that basis, we prove that there there is a real
number that is not in the range of the enumeration. There is nothing
used except first order logic applied to the axioms of set theory. To
reject the argument you have three choices:
1. Reject first order logic.
2. Reject at least one of the axioms of set theory used for the
argument.
3. Show a step in the argument that uses anything not authorized by
first order logic applied to the axioms of set theory..
And item 3 is not on the table, since there is no step in the argument
that is not authorized by first order logic applied to the axioms of
set theory.
MoeBlee
MoeBlee
> Define a real number r between 0 and 1. Denote
> r_n to be the nth digit of r. The digits of r are defined
> as follows:
>
> r_n = 5 if r_n = 4, otherwise r_n = 4.
That's nonsense and has no bearing on uncountability proofs.
MoeBlee
MoeBlee
I said no such thing by analogy or otherwise.
> Because we are using the verbage "ANY", we don't
> have to worry about special cases like when x = 0.
> That's how mathematicians work?
No, you've got it reversed. If we say 'any' to mean an arbitrary one,
then we don't assume that it falls under a special case or that it
doesn't fall under a special case. If there are special cases such that
the property does not hold of those objects that are a special case,
then the universal genearlization WON'T go through. It is the VERY
NATURE of the rule of universal generalization that it won't go through
if there is even a single special case that precludes the
generalization.
You seem completely unfamiliar with such basic principles of
mathematical reasoning. Why don't you read a book on the subject?
MoeBlee
Ross A. Finlayson
It helps to address all the "uncountability" proofs, or as I call them,
rrresults, at once, all three of them.
So I do.
Why doesn't anybody answer me?
They do.
Ross
MoeBlee
No, that's just nonsense. It is certainly not within the auspices of
the logic of such mathematical arguments as those showing the
uncountability of the reals.
MoeBlee
is just an example of the kind of argument that mathematical reasoning
does NOT allwo.
Alan Morgan
But if it's true for ANY list then it must be true for a specific
list. So if considering a single specific list shows a flaw then
perhaps that list doesn't really exist.
It's as if, after hearing the proof that sqrt(2) is irrational,
you reply "But what if I give you a and b, both integers, such
that a/b = sqrt(2)? Your argument is demolished!". Yes, if
you provide what the proof shows can't exist then the proof is
wrong. But you can't just assume that that thing exists.
Alan
--
Defendit numerus
MoeBlee
No, I refuted you directly in previous posts.
The problem here is very basic: You don't understand what a
mathematician means when he argues such as this: "Let x be an arbitrary
even number. [Then he proves some property P about x.] So, since x is
arbitrary, property P holds for all even numbers."
MoeBlee
Poker Joker
I did.
>> Because we are using the verbage "ANY", we don't
>> have to worry about special cases like when x = 0.
>> That's how mathematicians work?
>
> No, you've got it reversed.
No, I don't. You neglected to point out the difference in
the analogy.
I assume you might not understand analogy so I provided
a link to a good site for you to start learning:
http://en.wikipedia.org/wiki/Analogy
> If we say 'any' to mean an arbitrary one,
> then we don't assume that it falls under a special case or that it
> doesn't fall under a special case. If there are special cases such that
> the property does not hold of those objects that are a special case,
> then the universal genearlization WON'T go through. It is the VERY
> NATURE of the rule of universal generalization that it won't go through
> if there is even a single special case that precludes the
> generalization.
After you figure out the analogy, then maybe you should come back
and discuss this some more. It's quite obvious you need to learn
the analogy before you'll understand.
> You seem completely unfamiliar with such basic principles of
> mathematical reasoning. Why don't you read a book on the subject?
I'm sure from a person who has no understanding of what an
analogy is, my argument may be incomprehensible and therefore
beyond your grasp. I'm sorry about that, but I'm sure the link I
gave you will help you in the future.
Randy Poe
What in the world are you talking about?
When I define f:N->R, I am defining a function that takes naturals
as input and produces reals as output. How the heck does sqrt(x)
fail that test? What do you mean by "defining x in terms of the
set of all reals"? f maps naturals to reals. x is a natural.
- Randy
Poker Joker
There also needs to be assurance that the property P actually holds
for all x. Just because some people can't come to grips with the
fact that there are special cases to consider, doesn't mean they
aren't there. I understand FULLY your thought process. The
ONLY difference between your understanding of the proof and
mine is that I understand the special case and you don't.
Randy Poe
Why does step 2 have "questionable validity"?
> > > Under the most general assumption, we can't count out that
> > > R is f's image, so defining a real in terms of the image of
> > > f *MIGHT* be self-referential, and it certainly is if the image
> > > of f is R.
> >
> > What is the problem here?
>
> I assume you accept this proof that there are no complete lists
> of reals:
>
> Let r be a real number between 0 and 1. Let r_n denote the nth digit
> in r's decimal expansion. Let r_n = 5 if r_n = 4, otherwise let r_n = 4.
That doesn't make sense. You are saying that every digit of r
both is equal to 4 and is equal to 5.
Consider r = 0.00000000...
So you're saying the first digit of r is 4 because the first digit of
r isn't 4? What the hell are you talking about?
> r isn't on any list of reals. Therefore there isn't a complete list of
> reals.
That bears no resemblance at all to a proof.
- Randy
William Hughes
Hardly, but this "proof" does not reflect what is being done.
You start out with a set of real numbers A, with a certain
property, there is a surjective function from the natural
numbers to A. In other words A is a list of real numbers.
You define a process D(A) which gives you a real number.
D depends only on the fact that a surjective function
from the natural numbers exists, it does not depend on
any other property of A. Thus D can be applied to any
list . In particular, if we assume there is a list
containing all the real numbers, D can be applied to this
list. That this application will lead to a contradiction
does not change the fact that D can be applied to the
list. So step 2 is valid.
Of course, we need not make this assumption. In this case
the proof goes
1 Let A be a list
2 Use D to contruct a real number r
3 Show that r in not an element of A
4 Conclude that A does not contain all the
real numbers
Again, since D can be applied to any list
step 2 is valid.
- William Hughes
MoeBlee
> >> By analogy, what you're saying is:
> >>
> >> For ANY x
> >> there is a procedure to find a y such that x/y = 1.
> >
> > I said no such thing by analogy or otherwise.
>
> I did.
Right, so when you said "By analogy, What YOU're saying is", [all caps
added], the word 'your'e' does not belong there.
> >> Because we are using the verbage "ANY", we don't
> >> have to worry about special cases like when x = 0.
> >> That's how mathematicians work?
> >
> > No, you've got it reversed.
>
> No, I don't. You neglected to point out the difference in
> the analogy.
I did. There's no analogy with any signficant feature of what I said,
and the reason I say that is just as I posted to which you just
responded by saying I don't understand the analogy.
My points stand. You've not refuted uncountability proofs nor Arturo's
point to which you originally responded.
If you reject first order logic, then fine. No one says you have to
accept first order logic. Then maybe ou can tell us what system of
logic for mathematics you use instead.
MoeBlee
MoeBlee
> > The problem here is very basic: You don't understand what a
> > mathematician means when he argues such as this: "Let x be an arbitrary
> > even number. [Then he proves some property P about x.] So, since x is
> > arbitrary, property P holds for all even numbers."
>
> There also needs to be assurance that the property P actually holds
> for all x.
The property holds for all even numbers x since it hold of an ARBITARY
even number x. That is reasoning that is as valid now as it always has
been.
> Just because some people can't come to grips with the
> fact that there are special cases to consider, doesn't mean they
> aren't there.
If there are special cases that disqualify the generalization, then the
generalization will NOT be permitted by the rule of universal
generalization. That is the REASON the rule is drawn up the way it is.
Look, suppose x is an arbitrary even number and I prove that therefore
x has property P. I then conclude that all even numbers have property
P.
Then you say, "What if x is prime? What if x is greater than 100? What
if x is a successor of a prime? What if x is square? What if any number
of special cases hold?"
Then I say, as any mathematician would say, "If any of those special
cases disqualified some even number from having property P, then I
would not have been able to prove that P holds for an ARBITRARY even
number."
You REALLY don't understand that?
MoeBlee
Poker Joker
"Alan Morgan" <amo...@xenon.Stanford.EDU> wrote in message
news:efkegr$6d9$1...@xenon.Stanford.EDU...
>>if its true for ANY list, then it must be
>>true for a specific list. So if considering a single specific list
>>shows a flaw, then looking at ANY (ALL of them) list doesn't
>>help.
>
> But if it's true for ANY list then it must be true for a specific
> list. So if considering a single specific list shows a flaw then
> perhaps that list doesn't really exist.
That's true, but that's not the entire story.
Suppose I claim that I have a list that contains all the reals.
You claim you can take that list and construct a real not
on the list. You procede to show the construction. I would
claim that your construction is flawed because it is
self-referential, which it must be if I truly gave you a list of
all the reals. So in that *SPECIAL CASE*, unlike the
general case, your construction isn't valid. The only way
to eliminate that special case is to use what the
conclusion of the proof would be if you neglected the
special case.
> It's as if, after hearing the proof that sqrt(2) is irrational,
> you reply "But what if I give you a and b, both integers, such
> that a/b = sqrt(2)? Your argument is demolished!". Yes, if
> you provide what the proof shows can't exist then the proof is
> wrong. But you can't just assume that that thing exists.
No its not. Its like if you give me a proof that zero isn't a real
number:
For ANY x, there is procedure to construct a y, such that x = 1/y.
When x = 0, there is no y.
Therefore x is not a real number.
If everybody neglects the fact that the construction isn't valid
for x=0, then the proof is flawless.
Poker Joker
> Look, suppose x is an arbitrary even number and I prove that therefore
> x has property P. I then conclude that all even numbers have property
> P.
The property could be say, that x is prime (assuming you neglect the
special case of 2 and nobody notices like you've done below).
> Then you say, "What if x is prime? What if x is greater than 100? What
> if x is a successor of a prime? What if x is square? What if any number
> of special cases hold?"
>
> Then I say, as any mathematician would say, "If any of those special
> cases disqualified some even number from having property P, then I
> would not have been able to prove that P holds for an ARBITRARY even
> number."
>
> You REALLY don't understand that?
Yes, I understand. I also understand other things that you don't. I guess
we'll leave it at that.
Poker Joker
I understand everything you understand. I just understand more, and
that confuses you.
MoeBlee
> No its not. Its like if you give me a proof that zero isn't a real
> number:
>
> For ANY x, there is procedure to construct a y, such that x = 1/y.
> When x = 0, there is no y.
> Therefore x is not a real number.
>
> If everybody neglects the fact that the construction isn't valid
> for x=0, then the proof is flawless.
The point is that we are NOT be able to prove that for any x there is y
such that x=1/y.
Whatever we argue about x, if there is special case blocking the
generalization from x, then then our argument will NOT go through for
an arbitrary real number x.
If we prove that for any x there is a y such that x=1/y, then would
have had to use the assumption that x not= 0. So that would NOT be an
argument regarding an arbitrary real number x.
Look, here's what we have:
Mr. Mathematician says: Let p be an arbitrary prime number. [Insert
proof here that p has the property that there is a prime number greater
than p.] Therefore, for ALL prime numbers p, we have that p has the
property that there is a prime number greater than p.
Goober Boober says: But what if p is even? What if p is greater than
100? What if p is a successor of a prime? Indeed, what if p is such
that there is NOT a prime greater than p? Hey, Mr. Mathematician, you
didn't take into consideration all these special cases, did you? No,
you didn't. So your argument is invalid.
MoeBlee
MoeBlee
> "MoeBlee" <jazz...@hotmail.com> wrote in message
> news:1159579135.5...@m7g2000cwm.googlegroups.com...
>
> > Look, suppose x is an arbitrary even number and I prove that therefore
> > x has property P. I then conclude that all even numbers have property
> > P.
>
> The property could be say, that x is prime (assuming you neglect the
> special case of 2 and nobody notices like you've done below).
So what if x is prime? We proved that as long as x is an even number,
then it has property P, and we assumed nothing about x other than that
it is an even number. So whether x is prime or not prime, we proved
that x has property p, therefore all even numbers have property p.
> > Then you say, "What if x is prime? What if x is greater than 100? What
> > if x is a successor of a prime? What if x is square? What if any number
> > of special cases hold?"
> >
> > Then I say, as any mathematician would say, "If any of those special
> > cases disqualified some even number from having property P, then I
> > would not have been able to prove that P holds for an ARBITRARY even
> > number."
> >
> > You REALLY don't understand that?
>
> Yes, I understand. I also understand other things that you don't. I guess
> we'll leave it at that.
No, you don't understand, otherwise you would not be talking about x
being prime. And you would not object to the uncountability proof on
the basis that you are.
We prove for an ARBITRARY enumeration into the reals that that
arbitrary enumeration has the property of not have the set of real
numbers as its range. So we conclude that ALL enumerations into the
reals have the property of not having the set of real numbers as its
range. And we don't have consider YOUR objections that we may not have
considered special cases or that we missed some special case. We don't
have to consider such objections because, IF there were a special case
that blocks the generalization, then it would have blocked the proof
for an ARBITRARY enumeration into the reals. That principle of
inferring from the arbitrary to the general is just plain logic,
ancient or modern logic, it doesn' t matter. It's plain logic.
MoeBlee