> Consider the object S = {
> a
> ab
> abc
> abcd
> abcde
> ..
> }
>
> How many a's are there? oo
There is one string "a", there are infinitely many symbols a.
> Now consider an equivalent representation of S = {
> {
> a
> }
>
> {
> a
> ab
> }
>
> {
> a
> ab
> abc
> }
> ...
> }
>
> how many a's are there? oo
Agreed.
> Consider the object T = {
> {
> a
> }
>
> {
> b
> aa
> }
>
> {
> c
> bb
> aaa
> }
> ...
> }
>
>
> You can see the last displayed element is the mirror of this element from S?
> {
> a
> ab
> abc
> }
>
>
> How many a's are in T?
oo many symbols a.
> All of you say "finite", the answer is of course infinite.
What makes you think we all say finite? This would be a first.
> This is the induction you perform...
>
> x is a finite list
>
> x
> xx is a finite list
>
>
> x
> xx
> xxx is a finite list
>
> THEREFORE
> x
> xx
> xxx
> ... is a finite list
No.
The induction is:
x is a finite list of finite strings.
x
xx is a finite list of finite strings.
x
xx
xxx is a finite list of finite strings.
x
xx
xxx
... is an infinite list of finite strings.
> Particularly
>
> 3 is a rational
> 3.1 is a rational
> 3.14 is a rational
> 3.141 is a rational
> 3.1415 is a rational
>
> THEREFORE 3.14.. is a rational
3.14... is not a finite string, therefor it is not on the list, therefor
your conclusion is invalid.
> The 1, 2, 3 oo deduction actually is inducion if you can rearrange it.
>
> x is finite
>
> o
> oo
> oox
> ooxx is finite CASE n (o is some finite section of x's)
>
> IMPLIES
>
> o
> oo
> oox
> ooxx
> ooXXX is finite CASE n+1
>
> By induction
>
> x
> xx
> xxx
> ... is a finite list.
Induction makes statements about the elements on the list, not the list
itself. You can make an infinite sequence of finite lists, but none of
them will look like your result of induction.
> Be careful assigning Is_Finite property to infinite triangular lists.
>
> c
> bb
> aaa
> ...
>
> how many a's where there?
there are 3 a's visibile, unknown number on the rest of the list.
>
> how many rows are there here?
>
> 1234
> 1 a
> 2 ab
> 3 abc
> ...
>
> Infinite rows
>
> how many columns are there here?
> a
> ab
> abc
> ...
>
>
> its triangular, same as the number of rows,
> there are INFINITE colums
There are infinite columns, but only finitely many columns on any given
row. You appear to be mistaking properties of elements of your sequence
with properties of the sequence itself.
--
Will Twentyman
email: wtwentyman at copper dot net
> > How many a's are in T?
>
> oo many symbols a.
>
> > All of you say "finite", the answer is of course infinite.
>
> What makes you think we all say finite? This would be a first.
> > c
> > bb
> > aaa
> > ...
> >
> > how many a's were there?
>
> there are 3 a's visibile, unknown number on the rest of the list.
Going once, going twice! Lucky I tested you after you switch sides.
Herc
Do the members occupy/populate every column of the infinite columns?
Herc
Perhaps you are just getting more precise, or perhaps I am learning that
I need to be more precise when talking with you. In any case, the
c,bb,aaa, ... is poorly defined, since I have no clue what the next item
on the sequence is.
No single member occupies every column.
Every column has a member that extends into it.
The question isn't completely clear to me. Hopefully one of the above
addresses what you were asking.
Note: there is no infinitieth column, nor is there an infinitely wide
member.
x the 1st row is finite length
o
oo
oox
ooxx the nth row is finite length (o is some finite section of x's)
o
oo
oox
ooxx n
ooxxx if the nth row is finite length, the n+1th row is finite length
By induction
x
xx
xxx
...
Every row is finite length.
--------------------------------------------
Is that valid induction?
Herc
Yes.
--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
Yes, now how is that different from what I said?
Very valid, although your notation might be better expressed:
1: x
2: xx
3: xxx
...
n: x...xxx
...
Here's another example of infinity wierdness:
A set contains the sequence 0.3, 0.33, 0.333, ... and no other elements.
Is 1/3 in the set, or not?
The answer is no, as every element in the sequence can be
represented (10^N - 1) / (3 * 10^N), for any integer N > 0.
This is a strictly monotonic sequence, which has as its
limit 1/3 -- but for any element, I can find another,
greater element (namely, the next one), and 1/3 does not
equal any of the elements though the elements come
arbitrarily close:
1/3 = 10^N / (3 * 10^N) != (10^N - 1) / (3 * 10^N)
>
> Herc
>
--
#191, ewi...@earthlink.net
It's still legal to go .sigless.
Try again.
x the 1st column is finite length
o
oo
oox
ooxx the nth column is finite length (o is some finite section of x's)
o
oo
oox
ooxx n
ooxxx if the nth column is finite length, the n+1th column is finite length
By induction
x
xx
xxx
...
Every column is finite length.
--------------------------------------------
Is that valid induction?
Herc
Yes, that's perfectly valid induction. But try this one:
0.9 * 0.9 = 0.81
0.99 * 0.99 = 0.9801
0.999 * 0.999 = 0.998001
0.9999 * 0.9999 = 0.99980001
...
0.999...9 * 0.999...9 = 0.999...998000...001
With proper definition methods (e.g., the 8 is the N'th digit,
1 is the (2N)'th digit) and a rigorous method of
computing N+1 given N's computation, one can
jump from N to N+1, as required by weak induction.
One can therefore attempt to conclude (wrongly, of course) that
0.999... * 0.999... = 0.999...98000...01, where somehow the
'...' have become infinite digit sequences.
Welcome to Gary Denke Math. :-) Of course it turns out in
slightly more logical mathematics that
0.999... * 0.999... = 0.999... = 1.
where the 8 and its following 0's and the lone 1 have
vanished into infinity.
Or one can try to prove that 1/3 is arbitrarily close to
any element in the set 0.3, 0.33, 0.333, ... . Turns out
1/3 is indeed arbitrarily close -- but 1/3 is still not
actually in the set.
I defeated that proof anyway, by multiplying from the most significant
digit first, which is actually a fine way to do multiplication.
>
> Or one can try to prove that 1/3 is arbitrarily close to
> any element in the set 0.3, 0.33, 0.333, ... . Turns out
> 1/3 is indeed arbitrarily close -- but 1/3 is still not
> actually in the set.
>
Its not in any finite subset, but its in the set in its infinite form. You all
declare the *entire* anti diag number is not listed, you all put precedence
on the infinite length of the numbers over the infinite amount. All possible
finite *prefixes* are present, "but the whole *infinitely* long number isn't", why
do you conceptualise the numbers to all digits that can't be encapsulated, yet
the list is equally long and you encapsulate that? ****
0.3,
0.33,
0.333,
...
is itself equivalent to 0.33..
Take the diagonal, 0.33..
How many 3's are on the diagonal?
How many of those 3's are contained in members?
Take an infinte list of numbers of increasing length and matching prefixes.
3
3.1
3.14
3.141
3.1415
..
What amount of information is present in that infinite list?
How accurate does the list pinpoint pi? What's the diagonal of the set?
When you integrate a region of a curve, you use a formula that takes
the limit of the area of the regions as the width of the regions approaches
0 and the number of regions goes to infinity. If you segmented the
integration into n E N regions and did the calculation then you only have
an approximation to the area. But if you let it go to infinity you get a
precise answer, the area under the curve isn't [0,1) its 1. You've done
the calculation, there's no use going back to a finite model of the infinite
process and declaring "as the process goes to infinity the area approaches 1",
the area = 1. The infinite list of computable numbers is no different, stop
treating them like they're running on a 486, they are function values and
for this example 1/3 is reached. Be it a property of the set, the value is
encapsulated in the countable infinite list.
Herc
So 0.999... * 0.999... != 1? Color me confused.
>
>>
>> Or one can try to prove that 1/3 is arbitrarily close to
>> any element in the set 0.3, 0.33, 0.333, ... . Turns out
>> 1/3 is indeed arbitrarily close -- but 1/3 is still not
>> actually in the set.
>>
>
> Its not in any finite subset, but its in the set in its infinite form.
Sets don't have "infinite forms". Are you now wanting to approximate
a *set* using an infinite sequence of proper subsets? (An interesting
idea but that won't get much of anywhere either.)
> You all declare the *entire* anti diag number is not listed, you all
> put precedence on the infinite length of the numbers over the infinite
> amount. All possible finite *prefixes* are present,
> "but the whole *infinitely* long number isn't", why
> do you conceptualise the numbers to all digits that can't be
> encapsulated, yet the list is equally long and you encapsulate that? ****
Well, lessee. You define a list of rational prefixes of numbers,
then purport that that entire set covers R. You're right in
at least one sense; given any two elements in the set
K = {i / 10^j, i, j in J}
another element can be found between them, and therefore any number
in R can be approximated by a strictly monotonic sequence in K.
I leave the proof of these to the interested reader, but they're
fairly obvious.
K is closed under addition and subtraction.
K is closed under multiplication.
K is NOT closed under division. The closure of K is of course Q.
K is NOT closed under operations such as sqrt() or a^b.
The closure of K is, of course, R.
K has cardinality aleph-0.
K has at least one metric:
i / 10^j < k / 10^m if and only if i * 10^m < k * 10^j.
K is not equal to R (or even Q), except by stretching '=' a bit,
which is one of the things you're doing.
Or one can look at it this way. If one enumerates K', which
is defined as
K' = {0} union {i / 10^j, i,j in J, 0 < i < 10^j, j > 0, (i mod 10) != 0}
= K intersect [0,1)
one will get a set which you've seen before:
K' = {0, 0.1, 0.2, ..., 0.9, 0.01, 0.02, ..., 0.99,
0.001, 0.002, ..., 0.999, ...}
One possible diagonal number is 0.333.... (1/3), as it turns out.
But 1/3 is not in the set as it would require that
10^n / 3 be an integer, an absurdity -- even though
every finite prefix is in the set.
>
> 0.3,
> 0.33,
> 0.333,
> ...
>
> is itself equivalent to 0.33..
> Take the diagonal, 0.33..
>
> How many 3's are on the diagonal?
> How many of those 3's are contained in members?
'3' is a digit. Exactly what are you asking here?
As it is, turns out it's aleph-0 * (aleph-0 + 1) / 2 = aleph-0,
and this can be proven by a simple threading construct.
>
> Take an infinte list of numbers of increasing length and matching prefixes.
>
> 3
> 3.1
> 3.14
> 3.141
> 3.1415
> ..
>
> What amount of information is present in that infinite list?
> How accurate does the list pinpoint pi? What's the diagonal of the set?
It doesn't. It gets as close as desired, but it doesn't pinpoint pi.
No rational approximation method can.
>
> When you integrate a region of a curve, you use a formula that takes
> the limit of the area of the regions as the width of the regions approaches
> 0 and the number of regions goes to infinity.
Standard Riemannian integration, yes. Lebesgue integration takes
an entirely different path, and I'll have to restudy Rudin's
work to get it exactly right (for starters, he uses
uncountably infinite open covers).
> If you segmented the integration into n E N regions and did the
> calculation then you only have an approximation to the area.
> But if you let it go to infinity you get a precise answer,
> the area under the curve isn't [0,1) its 1. You've done
> the calculation, there's no use going back to a finite model
> of the infinite process and declaring "as the process goes to
> infinity the area approaches 1", the area = 1. The infinite
> list of computable numbers is no different, stop treating them
> like they're running on a 486, they are function values and
> for this example 1/3 is reached. Be it a property of the set,
> the value is encapsulated in the countable infinite list.
Is sqrt(2) rational, or not?
Your argument holds for any arbitrarily large set, not for an infinite set.
Pi is consisely represented here, its just a square on the number of digits.
Does this number contain 3-pi?
0.1141411415141591415921415926141592651415926..
Herc
pi - 3!
>
> 0.1141411415141591415921415926141592651415926..
>
> Herc
>
>
>
Not sure what you mean by "contain". If I'm reading
this correctly you are concatenating successively
longer prefixes of the number x = pi-3.
This is a semicomputable number, as pi is a semicomputable number.
Might as well ask whether this number contains pi-3:
0.51415926535...
(this is actually (pi-3)/10 + 1/2).
or this:
0.5154515559525655...
(interleaving with 5's)
or this:
0.104001000500009000002000000600000005....
(interleaving with successively longer strings of 0's).
Please clarify your definition of "contains" in this context.
No. You have not properly stated your concepts. If there is a finite
number of rows, the 1st column is finite. If there are a finite number
of rows, the nth row being finite implies the (n+1)th row is finite.
You conclusion switches to infinite rows, thus invalidating everything.
To put it another way, in the infinite rows case, where is the finite
column that will be used as your base case?
OK, color me slightly confused here.
The problem is similar to this one, AFAICT:
1 is finite.
If n is finite, so is n+1.
Therefore, every natural number is finite.
Or:
S_1 = {1} is a finite set.
If S_n is a finite set, then S_{n+1} = S_n union {n+1} is, too.
Therefore, all S_i are finite.
Perfectly valid weak induction in both cases. Of course,
one can get slightly squirrely here, if one is not very
careful, as one cannot thereby conclude that *every* subset
of N is therefore finite. The even numbers, in
particular, is an infinite set. I think this the pit
whereinto |-|erc might fall next, if he's not careful. :-)
|-|erc also seems to be of the opinion that one can somehow
identify the sup of a set with the set (this is fine), and
that therefore the sup is somehow in the set, closing it
(this is not). His simple case:
{0.3, 0.33, 0.333, ... }
does not include the sup 1/3, as is easily proven by computing the
terms explicitly (0.333...3 = (10^n - 1) / (3*10^n) for some n),
even though it includes every finite prefix.
The ellipsis, however, is a tricky beast, and you are correct
in that one has to be slightly careful in its application.
pretty much granted
>
> The problem is similar to this one, AFAICT:
>
> 1 is finite.
>
> If n is finite, so is n+1.
>
> Therefore, every natural number is finite.
>
more like this.
1 a finite amount of numbers
1, 2 a finite amount of numbers
1, 2, ...n has a finite amount of numbers
THEN
1, 2, ...n, n+1 has a finite amount of numbers
Therefore (for all numbers) there are a finite amount
Herc
Induction is not universal. There are some properties on which you can
do induction (for instance, x=0, x=x, x is finite, (x+1)+1=x+2 ), other
on which it does not make a sense to use induction. Logicists call them
"second order formulas", meaning that they are properties that depend on
some relation between the *sets* of elements, not only the elements
themselves. Think that indution can be applied just to properties that
can be expressed using the symbols
0,1,+,*,(,),=,[AND],[OR],[NOT],[FORALL],[EXISTS].
So "the set of numbers less than n contains a finite amount of numbers"
is true for any n, but the correct application of induction leads to
"for all x, the set of numbers below x is finite".
right! hence my subject line.
This is my interpretation of diagonalisation.
1
1 n
f(1) = (1,1) mod 2 + 1 forms a new number
1 2
1 n o
2 p q
f(1) = (1,1) mod 2 + 1
f(2) = (2,2) mod 2 + 1
f(1).f(2) forms a new number
---------------------
1 2 ... n
1 a b c d
2 e f g h
.. i j k l
n
for i = 1 to n
E.f(i) = (i,i) mod 2 + 1
forms a new number
IMPLIES
1 2 ... n n+1
1 a b c d m
2 e f g h n
.. i j k l o
n p q r s t
n + 1
for i = 1 to n + 1
E.f(i) = (i,i) mod 2 + 1
forms a new number
-------------------------
By induction, x e N f(x) = (x,x) mod 2 + 1 (in sequence) forms a new number
Herc
Based on the diagram, he appeared to make a jump from the finite case to
the infinite case. I may have misunderstood what he was trying to
represent with his final diagram, of course.
[valid induction deleted]
>
> Perfectly valid weak induction in both cases. Of course,
> one can get slightly squirrely here, if one is not very
> careful, as one cannot thereby conclude that *every* subset
> of N is therefore finite. The even numbers, in
> particular, is an infinite set. I think this the pit
> whereinto |-|erc might fall next, if he's not careful. :-)
>
> |-|erc also seems to be of the opinion that one can somehow
> identify the sup of a set with the set (this is fine), and
> that therefore the sup is somehow in the set, closing it
> (this is not).
That is the precise issue I was trying to forstall with my comment. It
*looked* like he was switching scenarios. Without a clear commentary, I
took the diagram at face-value as a changed scenario and declared it
invalid to switch cases.
> His simple case:
>
> {0.3, 0.33, 0.333, ... }
>
> does not include the sup 1/3, as is easily proven by computing the
> terms explicitly (0.333...3 = (10^n - 1) / (3*10^n) for some n),
> even though it includes every finite prefix.
>
> The ellipsis, however, is a tricky beast, and you are correct
> in that one has to be slightly careful in its application.
Or, at times, excruciatingly careful.
Induction does hold for second order formulas as they are usually
understood. I don't see what makes you think otherwise.
--
Aatu Koskensilta (aatu.kos...@xortec.fi)
"Wovon man nicht sprechen kann, daruber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
>"The Ghost In The Machine" <ew...@aurigae.athghost7038suus.net> wrote in
[snip]
>> Or one can try to prove that 1/3 is arbitrarily close to
>> any element in the set 0.3, 0.33, 0.333, ... . Turns out
>> 1/3 is indeed arbitrarily close -- but 1/3 is still not
>> actually in the set.
>>
>
>Its not in any finite subset, but its in the set in its infinite form.
That, sir, is gibberish. It either IS or IS NOT one of the members of the
infinite set. If it is not a member (which it isn't) then it is not "in" the
set in any useful sense, since for sets "is in" precisely means "is a member
of".
> You all declare the *entire* anti diag number is not listed,
Correct.
> you all put precedence on the infinite length of the numbers over the
> infinite amount.
(I infer you are attempting to refer to the infinite size of the set.)
> All possible finite *prefixes* are present,
Correct.
> "but the whole *infinitely* long number isn't",
Correct.
>why
>do you conceptualise the numbers to all digits that can't be encapsulated,
> yet the list is equally long and you encapsulate that? ****
Every member of the set has ("encapsulates" in your idiosyncratic term) a
*finite* count of "3"s; the set itself has ("encapsulates") an infinity of
members. This really really really isn't a problem.
Note that the list IS NOT "equally long" -- it is infinite, and each
particular sequence of "3"s is finite.
>0.3,
>0.33,
>0.333,
>...
>
>is itself equivalent to 0.33..
NO - IT - ISN'T. The set CONTAINS all finite sequences of "3"s, it is not
itself a sequence of "3"s (finite or infinite). They used to teach the
difference between sets and their members to *primary school* children for
goodness sake; it's hard to credit that you are too thick that to understand
it.
But your main problem is the Phillite Fallacy, wherein the infinite set of
natural numbers {0, 1, 2, ...} (for the above example, each number is the
count of "3"s in a particular sequence) somehow contains "infinity" as a "last
element". It doesn't, no matter how much you might wish it to.
that is the retort of a coward who fails to address the issue. you always carry
on the 'infinite form' of the diag number is not there, but when told the infinite
form of the set you break down.
xxx
xxx
xxx
THIS has a new diag element because the # of members are *limited.*
xxx
xxx
xxx
...
THIS has *unlimited members*, by defintion there is no missing sequence. you induced
your solution of hyper infinities going from finite case to infinite case.
Herc
I doubt this will help |-|erc's problem, but one can
define real numbers as Cauchy sequences; if one strongly
types the numbers, leading to representations as 1N, 2N,
-3J, 3J !=J 3N, 0J, and extend J to Q, leading to 3/1Q,
5/2Q, etc., one can then take Cauchy sequences in the usual
manner and define the number 0.333...R != 0.333...Q = 1/3Q
as the limit of the numbers in the Cauchy sequence S_Q =
{0.3Q, 0.33Q, 0.333Q, ...}, which of course is a subset
of Q. Since 0.333...R is a real it's not part of S_Q.
S_R can be generated from S_Q in a fairly obvious fashion
(in a pinch use a Cauchy sequence where every term is
the same!), but 0.333...R is not a member of either set
anyway, although one can make a case for 0.333...R being
associated with S_Q, as S_Q defines 0.333...R, but then,
so do a lot of other sequences, such as {1/4Q, 2/7Q, 3/10Q,
4/13Q, ...} and {10/31Q, 100/301Q, 1000/3001Q, ...}.
Most people don't bother. :-) (My background is in
computer programming, and it does make a difference in some
cases whether '1' is represented as a char, int, long,
longword array (arbitrary precision integer), floating
point, double-precision floating point, character string,
etc., so for me it's a fairly obvious if slightly useless
thing in theoretical math to worry about. :-) )
If one wants to get really weird one can throw set theory
into the mix, resulting in 0W = {}, 1W = {{}}, 2W = {{},{{}}}, etc.
Not sure what -1J would look like but this might be a weakly
typed system anyway, as 1J = 1W and -1J is merely "that entity
such that 1J + (-1J) = 0J" Ditto for 1/3Q = "that entity
which when multiplied by 3/1Q becomes 1/1Q".
The best I can suggest is that one can "officially"
define a real number by using its decimal expansion to
generate a Cauchy sequence (and other Cauchy sequences
can be transformed to such an "official" sequence, with
a little work). In this case, 0.333...R = {0.3Q, 0.33Q,
0.333Q, ...}, but 0.333...R is not contained in itself,
and no set containing all elements of 0.333...R will
automatically contain 0.333...R; it will merely contain
0.333...R as a subset.
(And then there's issues such as whether {0.9Q, 0.99Q, 0.999Q, ...}
and {1.0Q, 1.00Q, 1.000Q, ...} define the same number. Ow,
my brain. :-) But this is a side jaunt.)
The question might be whether the list 0.3Q, 0.33Q, 0.333Q, ...
is equal to 0.333...R, defines 0.333...R (which is
possible), or whether any set containing the list 0.3Q, 0.33Q, ....
will automatically contain 0.333...R (no), or whether
|-|erc's talking about 0.333...R (probably) or 0.333...Q (not likely).
But one can only accept |-|erc's |-|ypothesis if one is excessively
(or deliberately!) sloppy about dealing with decimal expansions.
I'll admit to some curiosity as to whether the |-|ypothesis is
equivalent to requiring that all sets be closed but it would have
to be properly formulated first.
If a set S contains the subset {0.3, 0.33, 0.333, ... }
does it also automatically contain 1/3?
Did you really want to rewrite a good chunk of set theory?
>
> xxx
> xxx
> xxx
> THIS has a new diag element because the # of members are *limited.*
>
> xxx
> xxx
> xxx
> ...
>
> THIS has *unlimited members*, by defintion there is no missing
> sequence. you induced your solution of hyper infinities going
> from finite case to infinite case.
If a set S is defined {0, 0.1, 0.2, ... 0.9, 0.01, 0.02, ..., 0.99,
0.001, 0.002, ... 0.999, ...} does it automatically contain
the entire interval [0,1)? Does it contain 1 as well, making
the interval [0,1]? Does it contain any irrational numbers
at all?
(Does anyone have a name for this set? Presumably someone's
fiddled with this before.)
1/3 turns out to be a perfectly good antidiagonal for
this set, as the first digit -- the only one in risk of
colliding with anything nonzero -- is 1, not equal to 3.
However, the set S does contain all finite prefixes,
by construction.
>"Barb Knox" <s...@sig.below> wrote in
>> "|-|erc" <got...@beauty.com> wrote:
>>
>> >"The Ghost In The Machine" <ew...@aurigae.athghost7038suus.net> wrote in
>> [snip]
>> >> Or one can try to prove that 1/3 is arbitrarily close to
>> >> any element in the set 0.3, 0.33, 0.333, ... . Turns out
>> >> 1/3 is indeed arbitrarily close -- but 1/3 is still not
>> >> actually in the set.
>> >>
>> >
>> >Its not in any finite subset, but its in the set in its infinite form.
>>
>> That, sir, is gibberish. It either IS or IS NOT one of the members of the
>> infinite set. If it is not a member (which it isn't) then it is not "in"
>> the
>
>
>that is the retort of a coward who fails to address the issue. you always
>carry on the 'infinite form' of the diag number is not there, but when
>told the infinite form of the set you break down.
[snip]
Ah, I think I see the rule you're wanting to introduce: If one allows an
infinite list of sequences, then one MUST allow the individual sequences in
the list to be infinite themselves.
Trying to talk mathematics with you can be an intellectually interesting
exercise, rather akin to a chess player being introduced to fairy chess. So,
given your above non-standard "fairy maths" rule about infinite lists of
sequences, here's another go at showing that the infinite sequence .333... is
legitimately (by your rules) not in the infinite list .3, .33, .333, ...
First, let every member of the infinite list be an *infinite* sequence:
. 3 0 0 0 0 0 ...
. 3 3 0 0 0 0 ...
. 3 3 3 0 0 0 ...
. 3 3 3 3 0 0 ...
...
That is, the i'th sequence in the list consists of i 3s followed by an
infinite tail of 0s. Clearly, for every i, the i-digit prefix of 333...
occurs in the list (as the i'th member in fact). However, EVERY sequence in
the list has an infinite tail of 0s, which 333... conspicuously lacks.
Therefore 333... can not be in the list.
So, do you have a further fairy maths rule to exclude this construction? Or
do you now see that 333... is not in the list? Or maybe you'll just change
the subject.
Very pretty. I for one would not have thought of that approach. :-)
It neatly negates the "interval specification" argument: briefly,
a finite digit expansion .d_1 d_2 ... d_n actually specifies any
number between the range .d_1 d_2 ... (d_n - 1) 500...
and .d_1 d_2 ... d_n 500..., exclusive, or, put another way, if
r_basis = .d_1 d_2 ... d_n, then the actual real number r is
somewhere in the interval (r_basis - 10^(-n)/2, r_basis + 10^(-n)/2).
An infinite decimal expansion, however, does not have this problem.
So...|-|erc...is 1/3 in that set, or not? :-)
> "Barb Knox" <s...@sig.below> wrote in
>
>> "|-|erc" <got...@beauty.com> wrote:
>>
>>
>>>"The Ghost In The Machine" <ew...@aurigae.athghost7038suus.net> wrote in
>>
>>[snip]
>>
>>>>Or one can try to prove that 1/3 is arbitrarily close to
>>>>any element in the set 0.3, 0.33, 0.333, ... . Turns out
>>>>1/3 is indeed arbitrarily close -- but 1/3 is still not
>>>>actually in the set.
>>>>
>>>
>>>Its not in any finite subset, but its in the set in its infinite form.
>>
>>That, sir, is gibberish. It either IS or IS NOT one of the members of the
>>infinite set. If it is not a member (which it isn't) then it is not "in" the
>
> that is the retort of a coward who fails to address the issue.
No, that is the reply of someone who understands the difference between
a member of a set and the limit of the sequence of the members of the
set written in a particular order. You identify the set with the limit
and consider the set to *be* a number. No one else does.
> you always carry
> on the 'infinite form' of the diag number is not there, but when told the infinite
> form of the set you break down.
>
> xxx
> xxx
> xxx
> THIS has a new diag element because the # of members are *limited.*
>
> xxx
> xxx
> xxx
> ...
>
> THIS has *unlimited members*, by defintion there is no missing sequence. you induced
> your solution of hyper infinities going from finite case to infinite case.
Having unlimited members does not mean something is missing.
Because I'm the only person here who knows how to program with infinite streams
and lazy parameter evaluation.
this data structure in a computer, which is quite possible :
{0.3,
0.33,
0.333,
...}
and this data structure in a computer,
{0.3333...}
are one and the same. That's how functional programs interpret an infinite list,
they segment it into increasing portions and apply functions to those.
THE SAME
There's a function to convert between the 2 formats, no data is lost, each function
has a symmetrical function to get the infinite stream in the processing paradigm of your choice.
Each of them contains and infinite number of 3s in sequence. You can disprove this
by 1 and 1 method only, by telling me the maximum number, if you CANNOT tell
me the maximum number of 3s in sequence in that list, then its NOT FINITE IS IT?
x
x
xx
x
xx
xxx
This sequence has a very simple property, the number of rows and columns
of each element are the same. If one is infinite, then so is the other.
3
3.1
3.14
..
that is pi, to a computer its pi, to cantorians its a indexed negative property matrix whos
only interpretation is rowwise extrapolation.
Herc
OK. But these "data structures" are no longer mathematical sets.
> That's how functional programs interpret an infinite list,
> they segment it into increasing portions and apply functions to those.
>
> THE SAME
>
> There's a function to convert between the 2 formats,
No need to have such; they're the same, remember? One might
have two functions to OUTPUT the conscars in different
formats, but that's about it.
> no data is lost,
Data is *always* lost when outputting an infinite list.
The '...' is for that express purpose. Suppose the number
was in fact 0.333339342984376536...? Suppose the number
was 0.33333 + pi / 100000? Suppose the number was
something completely different?
Would the outputs change? If so, how?
> each function
> has a symmetrical function to get the infinite stream in the
> processing paradigm of your choice.
>
> Each of them contains and infinite number of 3s in sequence.
> You can disprove this by 1 and 1 method only, by telling me
> the maximum number, if you CANNOT tell me the maximum number
> of 3s in sequence in that list, then its NOT FINITE IS IT?
>
> x
>
> x
> xx
>
> x
> xx
> xxx
>
>
> This sequence has a very simple property, the number of rows and columns
> of each element are the same. If one is infinite, then so is the other.
>
> 3
> 3.1
> 3.14
> ..
>
>
> that is pi, to a computer its pi, to cantorians its a indexed
> negative property matrix whos only interpretation is rowwise
> extrapolation.
That could equally well be the number 3 + 14/99. Data has been lost.
I meant
{
int(pi()),
int(pi()*10)/10 ,
int(pi()*100)/100,
...
}
or are you calling 3.1415... not pi either? data has been lost?
Herc
> "Will Twentyman" <wtwen...@read.my.sig> wrote > >>
>
>>>>>>Or one can try to prove that 1/3 is arbitrarily close to
>>>>>>any element in the set 0.3, 0.33, 0.333, ... . Turns out
>>>>>>1/3 is indeed arbitrarily close -- but 1/3 is still not
>>>>>>actually in the set.
>>>>>>
>>>>>
>>>>>Its not in any finite subset, but its in the set in its infinite form.
>>>>
>>>>That, sir, is gibberish. It either IS or IS NOT one of the members of the
>>>>infinite set. If it is not a member (which it isn't) then it is not "in" the
>>>
>>>that is the retort of a coward who fails to address the issue.
>>
>>No, that is the reply of someone who understands the difference between
>>a member of a set and the limit of the sequence of the members of the
>>set written in a particular order. You identify the set with the limit
>>and consider the set to *be* a number. No one else does.
>>
>
>
> Because I'm the only person here who knows how to program with infinite streams
> and lazy parameter evaluation.
We're not talking about programming, we're talking about sets. No lazy
parameter evaluation, no streams (infinite or otherwise). You are
*interpretting* a set to be a number. That does not make the set
*equal* the number.
My recommendation is that you leave all programming concepts at the door
when you post to sci.logic. You are allowing how you write programs
dictate how you think about the math/logic being discussed. I know of
at least two programming instructors who have said, "Programming is
math." They never once said, "Math is programming." It is a subset
relationship. You cannot use properties of the integers to reliable
tell you about properties of the complex numbers.
> this data structure in a computer, which is quite possible :
>
> {0.3,
> 0.33,
> 0.333,
> ...}
>
> and this data structure in a computer,
>
> {0.3333...}
>
> are one and the same. That's how functional programs interpret an infinite list,
> they segment it into increasing portions and apply functions to those.
I doubt that you have ever encountered a truly infinite list or data
structure on a computer. Finite representations of infinite lists, but
not infinite lists. If you have, can I buy the computer from you? My
computer keeps running out of RAM and hard drive space.
> THE SAME
>
> There's a function to convert between the 2 formats, no data is lost, each function
> has a symmetrical function to get the infinite stream in the processing paradigm of your choice.
Here's the key point: you have to *convert* between the formats. If
they were identical, there would be no conversion necessary.
> Each of them contains and infinite number of 3s in sequence. You can disprove this
> by 1 and 1 method only, by telling me the maximum number, if you CANNOT tell
> me the maximum number of 3s in sequence in that list, then its NOT FINITE IS IT?
>
> x
>
> x
> xx
>
> x
> xx
> xxx
I'll say it one more time. A list is not the same as the limit of the
elements on the list. That is why we have the limit function/operator.
> This sequence has a very simple property, the number of rows and columns
> of each element are the same. If one is infinite, then so is the other.
This is pure nonsense. Consider the following sequence:
1,2,1,2,1,2,1,2,...
> 3
> 3.1
> 3.14
> ..
>
> that is pi, to a computer its pi, to cantorians its a indexed negative property matrix whos
> only interpretation is rowwise extrapolation.
I thought your list was 3 14/99. I must have misread it. Hint: a
computer has no clue what pi is. It just works with whatever you tell
it to work with. A computer does not assign meaning to what it works
with. I can define PI to be 2.71828182846 and E to be 3.14159265359 and
it will oblige me. Another programmer reading my code may have issues
with it, but the computer will not care.
Classic!
>
> I'll say it one more time. A list is not the same as the limit of the
> elements on the list. That is why we have the limit function/operator.
Is this induction valid?
<
<1>
>
the sequence <2> is not in the list
:: the list is missing a finite sequence of digits
<
<12>
<34>
>
the sequence <25> is not in the list
:: the list is missing a finite sequence of digits
<
<123>
<456>
<789>
>
the sequence <260> is not in the list (ignore recurring 9s)
:: the list is missing a finite sequence of digits
Case n
<
<xx..3>
<xx..6>
..
<xx..9>
>
the sequence <yy..0> (the mod_diag) is not in the list
:: the list is missing a finite sequence of digits
IMPLIES
Case n+1
<
<xx..3a>
<xx..6b>
..
<xx..9c>
<xx..0d>
>
the sequence <yy..0e> (the mod_diag) is not in the list
:: the list is missing a finite sequence of digits
THERFORE
<
<x>
<xx>
...
>
This list is missing a finite sequence of digits.
___________________________________________
Is this valid?
0
0.1
0.12
This sequence contains the diagonal, 0.12
0
0.3
This sequence contains the diagonal, 0.3
0
0.3
0.33
0.333
0.3333
0.33333
This sequence contains the diagonal, 0.33333
3
3.1
3.14
3.141
This sequence contains the diagonal, 3.141
0
0.3
0.33
0.333
..
This sequence contains the diagonal, 0.33..
Herc
> "Will Twentyman" <wtwen...@read.my.sig> wrote
>
>>My recommendation is that you leave all programming concepts at the door
>>when you post to sci.logic.
>
> Classic!
Yet it would probably help clear up your thinking about a wide range of
matters. Math is concerned with *definitions*, not *interpretations*.
>>I'll say it one more time. A list is not the same as the limit of the
>>elements on the list. That is why we have the limit function/operator.
>
>
>
> Is this induction valid?
>
> <
> <1>
>
> the sequence <2> is not in the list
> :: the list is missing a finite sequence of digits
>
> <
> <12>
> <34>
>
> the sequence <25> is not in the list
> :: the list is missing a finite sequence of digits
I'm not sure where the 25 came from.
>
> <
> <123>
> <456>
> <789>
>
> the sequence <260> is not in the list (ignore recurring 9s)
> :: the list is missing a finite sequence of digits
nor the 260
>
> Case n
> <
> <xx..3>
> <xx..6>
> ..
> <xx..9>
>
> the sequence <yy..0> (the mod_diag) is not in the list
> :: the list is missing a finite sequence of digits
nor am I sure what you mean by "the mod_diag"
>
> IMPLIES
>
> Case n+1
> <
> <xx..3a>
> <xx..6b>
> ..
> <xx..9c>
> <xx..0d>
>
> the sequence <yy..0e> (the mod_diag) is not in the list
> :: the list is missing a finite sequence of digits
>
>
> THERFORE
> <
> <x>
> <xx>
> ...
>
> This list is missing a finite sequence of digits.
I don't understand your argument. It may be valid, or not. But right
now it just doesn't make sense to me.
> ___________________________________________
>
> Is this valid?
>
> 0
> 0.1
> 0.12
> This sequence contains the diagonal, 0.12
Correct. The last term is 0.12.
>
> 0
> 0.3
> This sequence contains the diagonal, 0.3
Correct. The last term is 0.3.
>
> 0
> 0.3
> 0.33
> 0.333
> 0.3333
> 0.33333
> This sequence contains the diagonal, 0.33333
Correct. The last term is 0.33333.
>
>
> 3
> 3.1
> 3.14
> 3.141
> This sequence contains the diagonal, 3.141
Correct. The last term is 3.141.
>
> 0
> 0.3
> 0.33
> 0.333
> ..
>
> This sequence contains the diagonal, 0.33..
Incorrect. This sequence does not contain a term with an infinite
decimal expansion. It may represent it, but it does not contain it.
Notice the change from the above cases. There is *no* last term.
Diag is <14> I added 1 to each digit, modified_diag = <25>
fairly standard operation
So you conclude, sequences of the form
<1>
<12>
<123>
don't contain the diagonal?
Does this sequence have a diagonal?
0
0.3
0.33
0.333
..
Herc
>Is this valid?
NO.
[snip]
>0
>0.3
>0.33
>0.333
>..
>
>This sequence contains the diagonal, 0.33..
As I posted out before, which you chose not to respond to:
Ah, I think I see the rule you're wanting to introduce: If one allows an
infinite list of sequences, then one MUST allow the individual sequences in
the list to be infinite themselves.
Trying to talk mathematics with you can be an intellectually interesting
exercise, rather akin to a chess player being introduced to fairy chess. So,
given your above non-standard "fairy maths" rule about infinite lists of
sequences, here's another go at showing that the infinite sequence .333... is
legitimately (by your rules) not in the infinite list .3, .33, .333, ...
First, let every member of the infinite list be an *infinite* sequence:
. 3 0 0 0 0 0 ...
. 3 3 0 0 0 0 ...
. 3 3 3 0 0 0 ...
. 3 3 3 3 0 0 ...
...
That is, the i'th sequence in the list consists of i 3s followed by an
infinite tail of 0s. Clearly, for every i, the i-digit prefix of 333...
occurs in the list (as the i'th member in fact). However, EVERY sequence in
the list has an infinite tail of 0s, which 333... conspicuously lacks.
Therefore 333... can not be in the list.
So, do you have a further fairy maths rule to exclude this construction? Or
do you now see that 333... is not in the list? Or maybe you'll just change
the subject.
--
so this is false?
<unsnip>
THERFORE
<
<x>
<xx>
...
>
This list is missing a finite sequence of digits.
>
let me see...
0.333333333333333 infinite number of 3s 3333333330000000 infinite number of 0s 000000
= 0.3..
Just check that on my infinite precision computer, yep that's fine!
Herc
> "Will Twentyman" <wtwen...@read.my.sig> wrote
It *has* a diagonal. It does not *contain* the diagonal. The choice of
verb makes a difference.
>
>
> So you conclude, sequences of the form
>
> <1>
> <12>
> <123>
>
> don't contain the diagonal?
Both the above and below have a diagonal. The one above also contains
its diagonal, while the one below does not.
> Does this sequence have a diagonal?
> 0
> 0.3
> 0.33
> 0.333
> ..
>
> Herc
They both have a diagonal, they do not both have the diagonal as one of
the terms of the sequence ("contain the diagonal").
you interpret every verb as "member_of".. draw, mark, scribe, plot the infinite set.
>
> They both have a diagonal, they do not both have the diagonal as one of
> the terms of the sequence ("contain the diagonal").
>
assume I have infinite memory. That's what countable infinite set is right?
you don't seem to get that far, you call it countable infinity but you always
put a bound on it, now actually give the computer INFINITE memory.
the computer represents an infinite set.
3
3.1
3.14
..
that diagonal number is represented inside the computer right?
Herc
> "Will Twentyman" <wtwen...@read.my.sig> wrote
>
>>It *has* a diagonal. It does not *contain* the diagonal. The choice of
>>verb makes a difference.
>
>
> you interpret every verb as "member_of".. draw, mark, scribe, plot the infinite set.
No. "has" suggests a property of the sequence. "contain" suggests
membership. "draw, mark, scribe" are unclear to me at best, and you've
never defined what you mean in a way that is meaningful other than as
"member of". I know how to plot members. I do not know how to plot a set.
>>They both have a diagonal, they do not both have the diagonal as one of
>>the terms of the sequence ("contain the diagonal").
>>
>
>
> assume I have infinite memory. That's what countable infinite set is right?
> you don't seem to get that far, you call it countable infinity but you always
> put a bound on it, now actually give the computer INFINITE memory.
>
> the computer represents an infinite set.
>
> 3
> 3.1
> 3.14
> ..
>
> that diagonal number is represented inside the computer right?
It is represented, but it is not a member of the sequence. The problem
as I see it is as follows: you appear to want membership and
representation to be interchangeable. They are not. Moreover, appear
to want to talk about properties of the representation as if that told
us properties of any given member.
Talking with you about math feels like trying to nail silly putty to a
wall. As I try to be precise (though sometimes with errors), I get the
impression that you try to avoid being precise. You seem to talk about
generalities or concepts, rather than specific details.
Try this: what does it mean for a sequence to represent its diagonal?
by "maths" you mean "something you read" here
> wall. As I try to be precise (though sometimes with errors), I get the
> impression that you try to avoid being precise. You seem to talk about
> generalities or concepts, rather than specific details.
>
> Try this: what does it mean for a sequence to represent its diagonal?
>
3
3.1
3.14
3.141
the diagonal is 3.141
All sets of this incremental type, every sequence of digits on the diagonal is present in the set
as a member also.
So if *some* diagonal on the set of computable numbers form the anti_diag,
every sequence of digits in anti_diag is present in members of the set.
Say this is the computables list :
0.1234
0.2111
0.1211
0.1121
The diagonal is 0.1111...
We know the set can't contain 0.2222...
but is can have all the finite prefixes.
0.2
0.22
0.222
0.2222
..
the diagonal of this subset of the computable numbers is anti_diag hence
there is no finite sequence of digits of anti_diag not present on the list
of computable numbers.
I tried to show consise mathematics with this conjecture, you just ignored it.
Case n+1
<
<xx..3a>
<xx..6b>
..
<xx..9c>
<xx..0d>
>
the sequence <yy..0e> (the mod_diag) is not in the list
:: the list is missing a finite sequence of digits
If you can comprehend my invented notation for select matrix elements you'll
understand my interpreation of what Cantor proved, because there's no
way on Earth computable numbers are missing any sequence of digits.
Herc
pi is (the symbolic representation of) a number. The
digit expansion is known to more than a billion places,
but the entire number can never be printed out as a decimal expansion;
there's not enough carbon in the Universe.
All you're showing is approximations therefor -- and you're
far from unique in that respect, since humans can't go
around writing pi either; it would take too long. :-)
One useful approximation is exactly representable in hex as
400921fb54442d18
which breaks down in the more or less usual IEEE754 fashion:
pi = 2^(0x400 - 0x3ff) * (1 + 0x921fb54442d18 / 0x10000000000000)
or more conventionally
2 * (1 + 2570638124657944 / 4503599627370496)
= 884279719003555/281474976710656
= 3.14159265358979311599796346854418516159057617187500000...
Since the original specification was 3.14159265358979323846
(M_PI on my machine), data has been lost again. This isn't
pi, but close enough to approximate the Earth's orbit to
a fraction of a millimeter. (The Earth's orbit is not
circular either, and that will lead to far larger errors
than substituting this number for pi; the eccentricity is
currently on the order of about 5 million km.)
Your set sequence above could be used as a representation of pi,
but it has the same problem as the simpler sequence
{0.3Q, 0.33Q, 0.333Q, ...} [*]; while one can use this to define
0.333...R (by a variant of Cauchy sequences), 0.333...R is not
*in* the set, merely defined by it.
>
> Herc
>
[*] 'Q' and 'R' in this case indicate strong number typing. Most
people don't bother. :-)
--
Does this set cover [0,1) ?
T = {0, 0.1, 0.2, ... 0.9, 0.01, 0.02, ,,, 0.99, 0.001, 0.002, ... 0.999, ... }
All finite decimal prefixes exist in that set, by construction.
Is 2/5 in that set? (Yes, it's the fifth element, 0.4.)
Is 1/3 in that set?
Is pi/4 = 0.78539816339744830962... in that set?
Is the number of times you've claimed 1/3 is in {0.3, 0.33, 0.333, ...}
divided by the total number of times you've posted to Usenet
in that set? :-)
If one simply rewrites the elements:
0.[0] 0 0 0 0 ....
0. 1 [0] 0 0 0 ....
0. 2 0 [0] 0 0 ....
0. 3 0 0 [0] 0 ....
0. 4 0 0 0 [0]....
...
as Barb did in her case, it turns out 1/3 = 0.33333... is also an
anti-diagonal for this set; this should be fairly easy to prove,
and obvious to at least see.
One can recursively define the set as follows:
T = {0, 0.1, 0.2, 0.3, 0.4, ..., 0.9}
union {x: x = y/10 + z/10, y in {0,1,2,3,4,5,6,7,8,9}, z in T}
and construct a somewhat Russelish subsequence:
T_1 = {0, 0.1, 0.2, 0.3, 0.4, ..., 0.9}
T_2 = T_1 union {x: x = y/10 + z/10, y in {0,1,2,3,4,5,6,7,8,9}, z in T_1}
T_3 = T_2 union {x: x = y/10 + z/10, y in {0,1,2,3,4,5,6,7,8,9}, z in T_2}
...
This is mildly interesting but IMO slightly pointless.
It's worth noting that Q01 = {set of all rational numbers in [0,1) }
satisfies the definition condition:
Q01 = {0, 0.1, 0.2, 0.3, 0.4, ..., 0.9}
union {x: x = y/10 + z/10, y in {0,1,2,3,4,5,6,7,8,9}, z in Q01}
so one has to be very careful here; strictly speaking T is the
intersection of all such sets.
Or one can attack the problem from this direction.
Given two infinite decimal expansions A and B, one can
define a procedure to check if they are equal, which halts
if they are not equal and doesn't halt if they are equal,
if the infinite expansions are truly infinite, as opposed
to being representable by a construction-procedure f(j)
which can be analyzed by a sophisticated program.
The flip side is, if this algorithm halts, the two numbers
are not equal. I can prove that t_n, where t_n is an
element in T, is such that t_n has a digit different from
1/3's expansion; therefore the two are not equal. In
fact, given the diagonalization above, that's absurdly
easy; t_n's n'th digit is always 0. QED.
> "Will Twentyman" <wtwen...@read.my.sig> wrote in > |-|erc wrote:
>
>>>>They both have a diagonal, they do not both have the diagonal as one of
>>>>the terms of the sequence ("contain the diagonal").
>>>
>>>assume I have infinite memory. That's what countable infinite set is right?
>>>you don't seem to get that far, you call it countable infinity but you always
>>>put a bound on it, now actually give the computer INFINITE memory.
>>>
>>>the computer represents an infinite set.
>>>
>>>3
>>>3.1
>>>3.14
>>>..
>>>
>>>that diagonal number is represented inside the computer right?
>>
>>It is represented, but it is not a member of the sequence. The problem
>>as I see it is as follows: you appear to want membership and
>>representation to be interchangeable. They are not. Moreover, appear
>>to want to talk about properties of the representation as if that told
>>us properties of any given member.
>>
>>Talking with you about math feels like trying to nail silly putty to a
>
> by "maths" you mean "something you read" here
Ok, talking with you about something you read feels like trying to nail
silly putty to a wall.
So what?
> I tried to show consise mathematics with this conjecture, you just ignored it.
No, I told you I didn't understand what you were saying. I now do
understand what you were saying, but had other points that seemed more
pressing.
> Case n+1
> <
> <xx..3a>
> <xx..6b>
> ..
> <xx..9c>
> <xx..0d>
>
> the sequence <yy..0e> (the mod_diag) is not in the list
> :: the list is missing a finite sequence of digits
>
>
> If you can comprehend my invented notation for select matrix elements you'll
> understand my interpreation of what Cantor proved, because there's no
> way on Earth computable numbers are missing any sequence of digits.
That means you do not understand what Cantor proved.
Effectively yes. If you cannot provide a number inbetween any of the above numbers
and the members then the *set difference* is 0.
I'll define set_minus.
SetA = {2, 3, 4}
NumA = 3.2
SetA SetMinus Numa = 0.2
Now what is T SetMinus Pi ? 0
memories...
> if the infinite expansions are truly infinite, as opposed
> to being representable by a construction-procedure f(j)
> which can be analyzed by a sophisticated program.
>
> The flip side is, if this algorithm halts, the two numbers
> are not equal. I can prove that t_n, where t_n is an
> element in T, is such that t_n has a digit different from
> 1/3's expansion; therefore the two are not equal. In
> fact, given the diagonalization above, that's absurdly
> easy; t_n's n'th digit is always 0. QED.
Herc
BARB WRITES
>there are an infinite set of digits of anti_diag not present on the list of
>computable numbers. Is that right?
Yes, if by "set" you actually mean "sequence";
So which is it,
all digit strings from anti_diag are on the list of computable numbers?
OR
there are digit strings from anti_diag which are not on the list of computable numbers?
>
> > I tried to show consise mathematics with this conjecture, you just ignored it.
>
> No, I told you I didn't understand what you were saying. I now do
> understand what you were saying, but had other points that seemed more
> pressing.
>
> > Case n+1
> > <
> > <xx..3a>
> > <xx..6b>
> > ..
> > <xx..9c>
> > <xx..0d>
> >
> > the sequence <yy..0e> (the mod_diag) is not in the list
> > :: the list is missing a finite sequence of digits
> >
> >
> > If you can comprehend my invented notation for select matrix elements you'll
> > understand my interpreation of what Cantor proved, because there's no
> > way on Earth computable numbers are missing any sequence of digits.
>
> That means you do not understand what Cantor proved.
>
It means you didn't think when you wrote this
You seem to talk about
generalities or concepts, rather than specific details.
If you want to use mathematics make the correct induction from my model,
its not difficult.
Herc
That's not much of a definition. Are you defining
A SetMinus n = min(a in A) abs(n - a)?
Or, if one prefers, the computational form:
func setMinus(Set<Number> a, Number b) returns Number
{
Number diff = positiveInfinity;
foreach el (a)
{
if(diff > abs(el - b)) diff = abs(el - b);
}
}
Or the recursive form:
func setMinus(Set<Number> a, Number b) returns Number
{
if(a == emptySet) return positiveInfinity;
else return min(abs(firstEl(a) - b), setMinus(a - {firstEl(a)}, b));
}
assuming Set<> operator-(Set<>, Set<>) is appropriately defined.
If so, that's fine. (The positiveInfinity is a bit of a kludge.)
>
> Now what is T SetMinus Pi ? 0
Actually, 2.14159265... but that's mostly because T only covers [0,1).
Had you asked about T SetMinus pi/4, or T SetMinus (pi - 3),
the value would be 0.
What is T SetMinus any r in [0,1)? 0.
I fail to see the point of defining this operation but it's not
inconsistent with current mathematics. However, it also doesn't
prove, denumerate, or show evidence for the denumerability of [0,1).
[rest snipped]
so if you agree {0.3, 0.33.., 0.333.., ... } setminus 1/3 = 0
then why do you disagree that the entire infinite set plots 1/3 on the number line?
I've been suggesting a *setplot* for months, used arduous language to describe
the infinite operation, not a bijection, is there no equivalent setplot?
Herc
Um.. I'm going to disagree on this one. T SetMinus r (where r is in
[0,1)) is 0 iff r is in T. Otherwise, it is undefined (There is no
minimum if r is not also in T.)
Sorry. :-) The definition of SetMinus can be roughly
translated as "find the nearest point in the set (in
this case, T) to the suggested real, and then return the
distance". Therefore, if r is in [0, 1], T SetMinus r is
0, as T is dense.
Not that it helps |-|erc's case all that much.
If you want to define 'T SetMinus r' as T - {r}, then the
operation may not be all that defined if r is not already
an element of T (I'd have to look but suspect T - {r} in
that case is simply T), and would return a different type
of result.
>
>>
>>I fail to see the point of defining this operation but it's not
>>inconsistent with current mathematics. However, it also doesn't
>>prove, denumerate, or show evidence for the denumerability of [0,1).
>
>>
>>[rest snipped]
>
I'm not sure what you mean by "plots". However, if we define
S = { (10^n - 1) / (3 * 10^n), n > 0, n in J }
= {0.3, 0.33, 0.333, ... }
then the modified set S(r), where
S(r) = union (s in S) (s - r/2, s + r/2)
(basically a union of intervals) for any r > 0, will of
course cover and contain 1/3.
Unfortunately, S itself will not.
> I've been suggesting a *setplot* for months, used arduous
> language to describe the infinite operation, not a bijection,
> is there no equivalent setplot?
T(r), defined analogously to S(r) for the set
T = { k / 10^n, n > 0, 0 <= k < 10^n, n,k in J }
= {0, 0.1, 0.2, ..., 0.9, 0.01, 0.02, ..., 0.99,
0.001, 0.002, ... 0.999, ... }
will cover [0,1] (in fact (-r, r+1) ) for any r > 0.
However, T will not -- it doesn't contain 1, 1/3, or pi/4,
for example (though T is dense and points can be found
that are arbitrarily close thereto).
>
> Herc
no its a numerical difference not an element elimination.
So what is {0.3, 0.33, 0.333, ... } setminus 1/3 ?
Herc
Actually, it's because T is dense that such a point doesn't exist.
let r be in [0, 1] but not in T. Assume that there exists an x in T
such that x is a nearest point in T to r. Because T is dense in [0,
1], and r is not equal to x, there exists a y in T such that r < y < x
(or r > y > x if r > x). This means that (y-r) < (x-r), i.e. that y
is closer to r than x. This contradict the assumption that x was the
closest element in T to r. QED.
>
>Not that it helps |-|erc's case all that much.
>
>If you want to define 'T SetMinus r' as T - {r}, then the
>operation may not be all that defined if r is not already
>an element of T (I'd have to look but suspect T - {r} in
>that case is simply T), and would return a different type
>of result.
T - {r} = T if r is not in T.
Zero, of course. Is that somehow supposed to mean that
1/3 is in that set?
No.
I did say "roughly", and admittedly one could define SetMinus in
two ways: find the nearest point, or find a limit given all points.
The latter always works.
It is, however, an interesting question -- and |-|erc has already
asked what SetMinus {0.3, 0.33, ...} - 1/3 is (the answer is
0 if one takes limits, undefined if one takes the first
definition, since a nearest point doesn't exist).
>
>>
>>Not that it helps |-|erc's case all that much.
>>
>>If you want to define 'T SetMinus r' as T - {r}, then the
>>operation may not be all that defined if r is not already
>>an element of T (I'd have to look but suspect T - {r} in
>>that case is simply T), and would return a different type
>>of result.
>
> T - {r} = T if r is not in T.
As I suspected, admittedly, after thinking about it.
>
>
>>
>>>
>>>>
>>>>I fail to see the point of defining this operation but it's not
>>>>inconsistent with current mathematics. However, it also doesn't
>>>>prove, denumerate, or show evidence for the denumerability of [0,1).
>>>
>>>>
>>>>[rest snipped]
>>>
>
No, this is nothing to do with set difference, its a
relational operator applying MINUS, *numerical* subtraction to each element.
select min(abs(element - 5) from tablenumbers
like you 1st put it
> >>>>A SetMinus n = min(a in A) abs(n - a)?
though SQL is not defined for infinite sets.
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Forall number less than all y, d (also less than all y) is bigger than or equal to them,
therefore forall number less than (1/3-0.3, 1/3-0.33, ...)
ie. 0, -0.5, -0.001, -10
d is the largest, i.e 0.
therefore
{0.3, 0.33,0.333...} setminus 1/3 = 0
Its basically a limit operator in logic like so :
x =
0.9
0.99
0.999
whats the 'maximum' without using limits?
forall numbers y bigger than every x
there is no number less than y and bigger than every x
therefore y = 1
any number less than 1 is bigger than some x.
any number bigger than 1 has another number inbetween all x & 1.
Herc
>"Barb Knox" <s...@sig.below> wrote
[snip]
Nope. Your fairy maths rule here says that the concatenation of two infinite
sequences results in a new infinite sequence. It doesn't.
To see this, let s_j be the j'th digit in your alleged infinite sequence
concat(333...,000...). Now clearly the index j for the first 3 digit is 1.
Since you claim this construction produces a sequence containing 0s, what is
the index j for the FIRST 0 digit?
As even you can see, there isn't one, since for EVERY natural number j, s_j =
3. Thus the 0s never actually appear in the concatenated sequence: your
concat(333...,000...) is equal to its first argument, 333... So, 333... does
NOT have any tail of 0s, unlike every sequence that is actually in the list.
Your move.
Define a numerical subtraction operation that ranges over all elements of a set.
Basically "how close" is a number to elements in a set.
{2, 3, 4} setminus 2.9 = 0.1
> >>>>A SetMinus n = min(a in A) abs(n - a)?
Definition and example {0.3, 0.33, 0.333, ...} setminus 1/3
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Forall numbers less than all y, d (also less than all y) is bigger than or equal to them,
therefore forall numbers less than (1/3-0.3, 1/3-0.33, ...)
ie. 0, -0.5, -0.001, -10
d is the largest, i.e 0.
therefore
{0.3, 0.33,0.333...} setminus 1/3 = 0
unless you can tell us a number between 1/3 and that set?
Herc
typo, any number less than 1 is *less* than some x.
Herc
[snip]
Don't try changing the subject. The fact is that every sequence in the list
has an infinite tail of 0s, and that 333... does not. Thus 333... is not in
the list. Do you accept that now?
Note that I am talking purely about sequences of digits; interpreting them as
real decimal fractions is neither necessary nor relevant to the point I am
trying to get you to see: that the sequence 333.... is nowhere in the given
list of sequences.
Makes no difference.
Define SetMinus(Set, n) as min (t in {abs(y - n): y in Set) t.
The answer is obviously 0. Ditto for sup {abs(y - n): y in Set}.
Define SetMinus(set, n) as
min(inf({y: y in Set and y >= n}) - n, n - sup({y: y in Set and y <= n})).
Still 0.
>
> select min(abs(element - 5) from tablenumbers
>
> like you 1st put it
>> >>>>A SetMinus n = min(a in A) abs(n - a)?
>
> though SQL is not defined for infinite sets.
SQL is perfectly defined for infinite sets (countable or no),
the same way it is defined for finite sets. It defines
a set of records.
There is no maximum. The sup is of course 1, which
is not in the set.
>
> forall numbers y bigger than every x
> there is no number less than y and bigger than every x
> therefore y = 1
>
> any number less than 1 is bigger than some x.
> any number bigger than 1 has another number inbetween all x & 1.
Correct, if I understand you correctly. Is 1 in the set, though? No.
>
> Herc
the whole is greater than the sum of parts
>
> Note that I am talking purely about sequences of digits; interpreting them as
> real decimal fractions is neither necessary nor relevant to the point I am
> trying to get you to see: that the sequence 333.... is nowhere in the given
> list of sequences.
>
its an in depth analysis of containment of 1/3 in the sequence
{0.3, 0.33, 0.333, ...} so its the same subject.
0.3000
0.33000
0.333000
The diagonal is 0.333.. so I don't accept that 0.3.. is not in the list, since
for every of the infinite number of digits in the diagonal there is a member
that contains that sequence, hence the infinite sequence of 3s is in the
list. There's no variables involved here the list is a fixed object it should
be trivial to analyse, how many 3s in sequence are there? ANSWER THE QUESTION
You could say its how big can a number be VS how many numbers are there
The answer to both questions is the same, UNLIMITED. There are
unlimited successive 3's. Your induction that it remains limited is wrong.
x this is a finite list
x
xx this is a finite list
x
xx
.. this is an infinite list
"list" is refering to either the rows or colums, its just a triangle.
|\
| \
| \
| \
| \
| \ i
| i \
| \
| \
here's a right angled infinite triangle.
the perpendicular height is infinity,
the hypotenuse is infinity,
the base isn't finite darling
I want my mega computer to calculate the circumference of a 10 meter circle.
pi=
{3
3.1
3.14
3.141
..
}
using this pi is it necessary to approximate to rationals?
The fact remains, the infinite sequence of 3s is on the list.
1
11
111
...
how many 1s are in sequence?
{1111111111111} is not missing from the sequence
{1111111111111111111} is not missing from the sequence
if sequences of digits are used to open safes, and my list opens the safe,
what's the conclusion?
Herc
That's the problem!
>
> Define SetMinus(Set, n) as min (t in {abs(y - n): y in Set) t.
> The answer is obviously 0. Ditto for sup {abs(y - n): y in Set}.
>
> Define SetMinus(set, n) as
>
> min(inf({y: y in Set and y >= n}) - n, n - sup({y: y in Set and y <= n})).
>
> Still 0.
>
...
>
> Correct, if I understand you correctly. Is 1 in the set, though? No.
so {0.3, 0.33, 0.333, ...} setminus 1/3 = 0.
then why can't I define a plot operation that works on infinite sets
like so :
input number
if SET setminus 1/3 =
AUNT SELMA HAS 1 HOUR TO LIVE!
ok, just figured out the final key in world media mind control.
input number
if SET setminus number = 0 then plot number.
that would plot both the members and the limits.
Herc
Eh? I fail to see a difficulty here.
SetMinus ( { (10^n - 1) / (3*10^n), n in J, n > 0} , 1/3 ) = 0.
However,
Inclusion ( 1/3, { (10^n - 1) / (3*10^n), n in J, n > 0} ) = false.
Despite your apparent thinking, this is not a contradiction.
>
>
>>
>> Define SetMinus(Set, n) as min (t in {abs(y - n): y in Set) t.
>> The answer is obviously 0. Ditto for sup {abs(y - n): y in Set}.
>>
>> Define SetMinus(set, n) as
>>
>> min(inf({y: y in Set and y >= n}) - n, n - sup({y: y in Set and y <= n})).
>>
>> Still 0.
>>
> ...
>>
>> Correct, if I understand you correctly. Is 1 in the set, though? No.
>
>
> so {0.3, 0.33, 0.333, ...} setminus 1/3 = 0.
>
> then why can't I define a plot operation that works on infinite sets
> like so :
"Plot" operation?
>
> input number
> if SET setminus 1/3 =
>
>
> AUNT SELMA HAS 1 HOUR TO LIVE!
>
> ok, just figured out the final key in world media mind control.
?!
>
>
> input number
> if SET setminus number = 0 then plot number.
>
> that would plot both the members and the limits.
If one interprets "plot" in a certain way, perhaps.
The set
T = {k/10^n, k and n in J, 0 <= k < 10^n, n > 0}
= {0.0, 0.1, 0.2, ... 0.9, 0.01, 0.02, ..., 0.99, ... }
is dense and can be used to "plot" every point on the
interval [0,1], if one interprets "plot" of a real r
to mean "cover with an open ball of width epsilon"
(r - epsilon/2, r + epsilon/2), for any fixed positive
epsilon. [*]
Plot(Set, epsilon) = union (y-epsilon/2, y+epsilon/2)
over all y in Set
Put another way, for any r in [0,1], I can find a subsequence
in T which gets arbitrarily close (T, after all, contains
all possible finite prefixes).
Is 1/3 in T? No, because 3 does not divide any power of 10.
The sequence {0.3, 0.33, 0.333, ... }, which is a proper subset
of T, however, can get as close as one wants (N = ceil(-log10(epsilon))).
How much clearer do I need to get on this explanation? :-)
(ObDumbQuestion: is there a name for the set T? Something
along the lines of Champernowne's constant (0.12345678910111213...)?)
>
> Herc
>
[*] epsilon can be replaced by a function e(n): J+ => R+;
the function of course defines a sequence. So long as
that sequence defines a series f(n) = sum(i=1 to n) e(i)
that diverges to positive infinity, one can use that as well.
An interesting exercise might be to characterize all such
sequences that can be used to cover the line. One rather
absurd sequence is
e(6) = 2, e(n) = 0 for n != 6
which simply plots one giant open interval on the point 0.5.
That's clear, so T marks 1/3 on the number line!
I didn't need epsilon, but the same effect,
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Forall numbers less than all y, d (also less than all y) is bigger than or equal to them,
therefore forall numbers less than (1/3-0.3, 1/3-0.33, ...)
ie. 0, -0.5, -0.001, -10
d is the largest, i.e 0.
therefore
{0.3, 0.33,0.333...} setminus 1/3 = 0
We could call T the 03033 set.
Herc
>"Barb Knox" <s...@sig.below> wrote >
>> Don't try changing the subject. The fact is that every sequence in the list
>> has an infinite tail of 0s, and that 333... does not. Thus 333... is not in
>> the list. Do you accept that now?
>
>the whole is greater than the sum of parts
Not in set theory it isn't. Have you ever heard of "extensionality"? I
didn't think so. Maybe if you read some decent introductory book on set
theory you wouldn't spout so much rubbish (at least on that subject).
>> Note that I am talking purely about sequences of digits; interpreting them as
>> real decimal fractions is neither necessary nor relevant to the point I am
>> trying to get you to see: that the sequence 333.... is nowhere in the given
>> list of sequences.
>>
>
>its an in depth analysis of containment of 1/3 in the sequence
>{0.3, 0.33, 0.333, ...} so its the same subject.
Whatever you are trying to mean by "containment", it is clearly different from
"being a member of the list". The subject now appears to be your aggressive
ignorance about basic mathematical notions and forms of reasoning. The good
news is that it's curable; the bad news is that you first have to recognise it
needs curing.
Just in case you feel like exposing yourself to some actual knowledge, have a
look at <http://planetmath.org/encyclopedia/AxiomOfExtensionality.html>.
[further Herculean rubbish snipped]
You sure get adamant I answer every question when you snip dozens of mine.
Its impossible to discuss anything with you Barb, I'm lucky to get a question
I want anwered in 6 months from you because you pop in with a text book quote,
degrade all the work, ignore everything then choof off until next month, usually
grumbling 'i'm right he's wrong'.
I did get this out of you last episode.
"Will Twentyman" <wtwen...@read.my.sig> wrote
> > but is can have all the finite prefixes.
> >
> > 0.2
> > 0.22
> > 0.222
> > 0.2222
> > ..
> >
> > the diagonal of this subset of the computable numbers is anti_diag hence
> > there is no finite sequence of digits of anti_diag not present on the list
> > of computable numbers.
>
> So what?
BARB WRITES
>there are an infinite set of digits of anti_diag not present on the list of
>computable numbers. Is that right?
Yes, if by "set" you actually mean "sequence";
So which is it,
all digit strings from anti_diag are on the list of computable numbers?
OR
there are digit strings from anti_diag which are not on the list of computable numbers?
********* answer goes here ********
Now check this if you can, I'm sure you could shorten my defn.
Definition and example {0.3, 0.33, 0.333, ...} setminus 1/3
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
Ed d e R, get the 'minimum' of y
Ab b e W,
Az z e T,
Av v e T,
d<=z, d is less than all y
b<=v, b is less than all y
Aa a e W,
d>=a
Because we are dealing with infinite sets, something you have no grasp on,
we alter the meaning of membership.
For a numerical system.
a e S <-> S setminus a = 0
I see no problem with this, set membership now includes the limits of infinite sets
using a simple logical framework to encapsulate infinite sets.
Does this set include a mapping to the digit 1?
9 = (0.9, 0.99, 0.999, 0.9999, ... }
9 setminus 0.9 = 0 0.9 is drawn on the number line
9 setminus 0.99 = 0 0.99 is drawn on the number line
9 setminus 0.98 = 0.01 0.98 is NOT drawn on the number line
9 setminus 1 = 0
Obviously 1 is drawn on the number line along with all the other members,
just by using a different interpretation of membership.
Is there a number between all the points drawn on the number line usually and 1?
no, it makes sense that the INFINITE set draws the number 1.
Herc
And all other numbers in the set [0,1] as well, if one
defines "mark" in a certain way.
>
> I didn't need epsilon, but the same effect,
You need epsilon. 1/3 is not in T. 1/3, however, is
arbitrarily close to certain points thereof.
>
> S setminus r = d
> <->
> Ax x e S, 0.3, 0.33, 0.333
> Ay y e T,
> abs(x - r) = y, 0.03.., 0.003.., 0.0003.. = y
>
> Ed d e R, get the 'minimum' of y
> Ab b e W,
> Az z e T,
> Av v e T,
> d<=z, d is less than all y
> b<=v, b is less than all y
>
> Aa a e W,
> d>=a
>
> Forall numbers less than all y, d (also less than all y) is bigger than or equal to them,
> therefore forall numbers less than (1/3-0.3, 1/3-0.33, ...)
> ie. 0, -0.5, -0.001, -10
> d is the largest, i.e 0.
> therefore
> {0.3, 0.33,0.333...} setminus 1/3 = 0
>
>
> We could call T the 03033 set.
T is what I've been calling {k/10^n, 0 <= k < 10^n, n > 0, k, n in J}.
It's basically the set of all finite decimal prefixes. This set
"covers" the interval [0,1], though only in the sense that it's dense.
>
> Herc
[snip for brevity]
> Does this set include a mapping to the digit 1?
>
> 9 = (0.9, 0.99, 0.999, 0.9999, ... }
>
> 9 setminus 0.9 = 0 0.9 is drawn on the number line
> 9 setminus 0.99 = 0 0.99 is drawn on the number line
> 9 setminus 0.98 = 0.01 0.98 is NOT drawn on the number line
> 9 setminus 1 = 0
>
> Obviously 1 is drawn on the number line along with all the other members,
> just by using a different interpretation of membership.
>
> Is there a number between all the points drawn on the number line
> usually and 1? no, it makes sense that the INFINITE set draws
> the number 1.
>
> Herc
>
So are you suggesting that the set
T = {k / 10^n, 0 <= k < 10^n, n > 0, n, k in J}
draws the entire line segment from 0 to 1, inclusive?
After all, T setminus r = 0 for any r, 0 <= r <= 1.
where is epsilon?
S setminus r = d
<->
Ax x e S, 0.3, 0.33, 0.333
Ay y e T,
abs(x - r) = y, {0.03.., 0.003.., 0.0003..} = T
Az e T
Aa, b, a<=d<=z
the only solution is d=0
d is (less than) the smallest difference
and the biggest of those.
say z =0
Aa, b, a<=d<=z
a <= z
so a <= 0
d >= a
so d >= 0
d <= z
so d <= 0
therefore d = 0
ITS LOGIC MAN, nothing to do with approximations arbitrarily close.
Herc
> > > the diagonal of this subset of the computable numbers is anti_diag hence
**********************************************************
> > > there is no finite sequence of digits of anti_diag not present on the list
> > > of computable numbers.
> >
> > So what?
>
> BARB WRITES
> >there are an infinite set of digits of anti_diag not present on the list of
> >computable numbers. Is that right?
> Yes, if by "set" you actually mean "sequence";
>
> So which is it,
> all digit strings from anti_diag are on the list of computable numbers?
> OR
> there are digit strings from anti_diag which are not on the list of computable numbers?
>
>
> ********* answer goes here ********
<unsnip question for Barb>
there is no finite sequence of digits of anti_diag not present on the list
of computable numbers.
Will : yes
there are an infinite set of digits of anti_diag not present on the list of
computable numbers.
Barb : Yes, if by "set" you actually mean "sequence"
You can see why I want this clarified.
Herc
Well, if you want, you can take the set T = {k/10^n, 0 <= k < 10^n, n > 0}
and use an epsilon function (which is simply a sequence generating
a series); the epsilon function's series need merely diverge to
infinity.
For example, epsilon(j) = 10^-5 / j will result in a complete cover.
>
> S setminus r = d
> <->
> Ax x e S, 0.3, 0.33, 0.333
> Ay y e T,
> abs(x - r) = y, {0.03.., 0.003.., 0.0003..} = T
>
> Az e T
> Aa, b, a<=d<=z
>
> the only solution is d=0
>
> d is (less than) the smallest difference
> and the biggest of those.
>
>
> say z =0
> Aa, b, a<=d<=z
>
> a <= z
> so a <= 0
>
> d >= a
> so d >= 0
>
> d <= z
> so d <= 0
>
> therefore d = 0
>
> ITS LOGIC MAN, nothing to do with approximations arbitrarily close.
That is correct. 1/3 is arbitrarily close to elements in S.
1/3 is not *in* S, however.
is {0.9, 0.99, 0.999, ...} = 1 ?
If so, where is 1 on the list?
what number is inbetween 0.999.. and 1?
what number is inbetween {0.9, 0.99, 0.999, ...} and 1?
the diagonal = 1
when a functional language reads from the keyboard it
just treats it like an infinitie stream.
if (first(keyboard(), 3) = "run") then run()
keyboard does not terminate by itself.
keyboard() = "runlistwin3543..."
To *interact* with the keyboard,
that is, read some input, do some processing, read some more input,
you have to FORMAT the infinite stream like so.
mapstream(keyboard()) =
<"r"
"ru"
"run"
"runl"
"runli"
"runlis"...>
Now a pure functional program can interpret an infinite stream and *interact*.
first(mapstream(keyboard()), 3) = "run"
Get the next command
first(cdr(cdr(cdr(mapstream(keyboard())))), 4) = "list"
functional programs can handle infinte streams.
the program
function nines ()
(cons 9 nines())
)
outputs 999999999999..
Its a dud program in 3GL, but in 4GL its a usable infinite data structure.
But nothing can process the infinite stream directly, it has to be FORMATTED
mapstream(nines()) =
9,
99,
999,
9999,
99999,
..
Now the computer can interpret 0.9... and get exactly the same results
as if it was processing the number 1.
Its sequences of this form :
e
ea
eac
each
eachl
eachle
eachlet
eachlett
eachlette
eachletter
eachletterl
eachletterli
eachletterlin
eachletterline
eachletterlines
eachletterlinesu
eachletterlinesup
that can be found in the computable numbers list as SUBSETS.
The diagonal of some of these subsets in anti_diag, Cantors missing number.
Therefore all digit sequences of irrational numbers are found on the
countable list.
Herc
> "Ellis Dees" <das...@my-deja.com> wrote
>
>>Herc,
>>
>>is {0.9, 0.99, 0.999, ...} = 1 ?
>>
>>If so, where is 1 on the list?
>
>
> what number is inbetween 0.999.. and 1?
They are both 1.
> what number is inbetween {0.9, 0.99, 0.999, ...} and 1?
Malformed question: what do you mean for something to be between a set
and a number?
You insist on doing comparisons between things where there is no defined
comparison operator.
--
Will Twentyman
email: wtwentyman at copper dot net
doesn't answer the question.
>
> > what number is inbetween {0.9, 0.99, 0.999, ...} and 1?
>
> Malformed question: what do you mean for something to be between a set
> and a number?
>
> You insist on doing comparisons between things where there is no defined
> comparison operator.
>
what number is between all members of the set and 1?
you guys can't see the forrest for the trees, you fail to recognise a set
contains a sequence of digits becuase of 'type' arguments applying over infinite objects.
Herc
So if I wanted to plot the SET onto the number line, I just do this.
if S setminus r = 0 then draw r.
Assume I can call (S setminus r) infinite times with every real.
Then I can do a 2nd plot drawing every member of the set, calling that infinite times,
once for every member.
What is the difference with the output, the markings onto the number line?
Herc
> "Will Twentyman" <wtwen...@read.my.sig> wrote >
> |-|erc wrote:
>
>>>"Ellis Dees" <das...@my-deja.com> wrote
>>>
>>>
>>>>Herc,
>>>>
>>>>is {0.9, 0.99, 0.999, ...} = 1 ?
>>>>
>>>>If so, where is 1 on the list?
>>>
>>>
>>>what number is inbetween 0.999.. and 1?
>>
>>They are both 1.
>
> doesn't answer the question.
There isn't one. What number is between 1 and 1? Nothing.
>>>what number is inbetween {0.9, 0.99, 0.999, ...} and 1?
>>
>>Malformed question: what do you mean for something to be between a set
>>and a number?
>>
>>You insist on doing comparisons between things where there is no defined
>>comparison operator.
>
> what number is between all members of the set and 1?
Again, nothing.
> you guys can't see the forrest for the trees, you fail to recognise a set
> contains a sequence of digits becuase of 'type' arguments applying over infinite objects.
You have to see the trees to see the forest. If you don't see trees,
it's something else.
U = { (10^n - 1) / 10^n, n in J, n > 0}.
Since 1 = 10^n / 10^n, having the value 1 in U would imply
that 10^n - 1 = 10^n for some integer n, which is absurd.
All points in [0,1] will be marked. You're basically plotting the
closure (?) of T, which is the given interval.
Ask Garry Denke. :-) As it is, any number
between 0.999... and 1 will have the same
expansion as 0.999..., which means that either
0.999... = 1 or one has a problem in uniquely
identifying precisely where 0.999... is.
Also, if x = 0.999... != 1, then 0.999... = 1 - d, and
10 * (x - 0.9) = 0.999... = 1 - 10 * d and
(x + 9) / 10 = 0.999... = 1 - 1/10 * d.
So what is d? If it's 0, no problem. Otherwise, 0.999...
is ambiguous again, and in fact one can make d as large
as one wishes (by using N * (x - (N-1)/N) and a large
enough N).
> what number is inbetween {0.9, 0.99, 0.999, ...} and 1?
>
That is a meaningless question, as phrased. If you want
to ask what is {0.9, 0.99, ...} SetMinus 1, then the
answer is of course 0.
Actually, there is no infinite stream as such, merely a procedural
representation that, if left to run without modification,
generates a Turing machine (or program) that loops without bound,
producing an endless stream -- but the program can also be modified
and optimized, with sufficiently complex algorithms.
function threeNines() ( first(nines(), 3))
for example, presumably outputs 9 9 9, since first() can
look at its first descriptor (which is *not* evaluated
prior to execution of first(), until a certain point)
and see that it's recursive.
>
> mapstream(nines()) =
> 9,
> 99,
> 999,
> 9999,
> 99999,
> ..
>
> Now the computer can interpret 0.9... and get exactly the same results
> as if it was processing the number 1.
It would take some doing. Of course, the real question is:
(isElementOf (/ pi 4) T)
where T is defined as per other posts. I'm not up on my LISP
but one might start defining T in the following fashion:
function T10set()
(list
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
... ?
(/ T10set() 10)
(+ (/ T10set() 10) 0.1)
(+ (/ T10set() 10) 0.2)
...
(+ (/ T10set() 10) 0.9)
)
This is not infinite tail recursion but might be
optimizable. I'm assuming here that (+) is defined on
a set and a scalar, adding the scalar to each element of
the set and constructing a new set, and (/) is defined on
a set and a scalar, dividing each element of the set by
the scalar and returning a new set.
pi, of course, can be computed in a number of interesting ways.
If you don't like pi, you can also use
(/ 1 3)
>
> Its sequences of this form :
> e
> ea
> eac
> each
> eachl
> eachle
> eachlet
> eachlett
> eachlette
> eachletter
> eachletterl
> eachletterli
> eachletterlin
> eachletterline
> eachletterlines
> eachletterlinesu
> eachletterlinesup
>
> that can be found in the computable numbers list as SUBSETS.
> The diagonal of some of these subsets in anti_diag, Cantors missing number.
> Therefore all digit sequences of irrational numbers are found on the
> countable list.
>
> Herc
>
>
>
Question 1
What number, if any, is between 0.99... and 1?
Question 2
What number, if any, is between all the numbers in {0.9, 0.99, ...} and 1?
What question do you think I'm trying to ask? What should I expect when
"forall x, trueformula(x) <-> hasproof(x)" disproves Godel statements so its not an
allowed formula.
pile = 0
start:
add sand to pile
goto start
is the pile of sand a mountain yet?
Herc
there is no finite sequence of digits of anti_diag not present on the list
of computable numbers.
Will : yes
there are an infinite set of digits of anti_diag not present on the list of
computable numbers.
Barb : Yes, if by "set" you actually mean "sequence"
what is the difference in the 2 plots of the set 03033?
Herc
No single number. For each member in the set, there is some number between it and 1.
{.0999...,.00999...,.000999...,.0000999...}
> you guys can't see the forrest for the trees, you fail to recognise a set
> contains a sequence of digits becuase of 'type' arguments applying over infinite objects.
So which member of that set actually contains the number 1?
Were you expecting "0.999...5"?
>
> Question 2
> What number, if any, is between all the numbers in {0.9, 0.99, ...} and 1?
>
Ditto.
>
> What question do you think I'm trying to ask? What should I expect when
> "forall x, trueformula(x) <-> hasproof(x)" disproves Godel statements
> so its not an allowed formula.
>
> pile = 0
> start:
> add sand to pile
> goto start
>
> is the pile of sand a mountain yet?
>
> Herc
> there is no finite sequence of digits of anti_diag not present on the list
> of computable numbers.
> Will : yes
>
> there are an infinite set of digits of anti_diag not present on
> the list of computable numbers.
> Barb : Yes, if by "set" you actually mean "sequence"
>
Actually, there are an infinite number of prefixes on your list.
Just not the final number.
Chalk it up to the weirdnesses of reals... :-)
Two plots?
The plot of the set of {0.3, 0.33, 0.333, ...} will cover 1/3, if
a plotted point is a non-zero width and/or one uses your
procedure to draw all real numbers such that 03033 SetMinus r = 0.
Since "and" equals "+" the number is "+"
The number "+" was adopted back in 1955.
You're welcome Herc.
Garry
> "Will Twentyman" <wtwen...@read.my.sig> wrote
>
>>>>>>is {0.9, 0.99, 0.999, ...} = 1 ?
>>>>>>
>>>>>>If so, where is 1 on the list?
>>>>>
>>>>>
>>>>>what number is inbetween 0.999.. and 1?
>>>>
>>>>They are both 1.
>>>
>>>doesn't answer the question.
>>
>>There isn't one. What number is between 1 and 1? Nothing.
>
>
> Question 1
> What number, if any, is between 0.99... and 1?
I've already answered this, I thought. Since 0.9999... = 1, there are
no numbers between them.
> Question 2
> What number, if any, is between all the numbers in {0.9, 0.99, ...} and 1?
A = {0.9, 0.99, ...}
B = {x | forall y in A, y < x < 1} = {}
None.
> What question do you think I'm trying to ask? What should I expect when
> "forall x, trueformula(x) <-> hasproof(x)" disproves Godel statements so its not an
> allowed formula.
Here's the thing: Sometimes my guess at what you're trying to ask is
incorrect. Other times, by guessing at what you're trying to ask, it
allows you to keep things a little bit fuzzy, so you can make arguments
that are not actually valid. One of my goals is to keep you fairly
precise so that you can't suddenly change draw conclusions based on
shifting your interpretation of what something means. I don't think
you'd do it to cause problems, but it can. I've made the same mistake
in the past, and gotten burned by it.
"forall x, trueformula(x) <-> hasproof(x)" is just a statement, which
may be true or false. It's an allowed formula, just not necessarily a
true formula or a provable formula.
> pile = 0
> start:
> add sand to pile
> goto start
>
> is the pile of sand a mountain yet?
At what iteration? As stated, this is only a set of instructions.
wrong question.
the topic is: does the infinite plot of points reach number 1 on the number line?
Herc
Is there an anti_diag number with this property?
0.123498765..
where 0.1234 is a prefix on CNL (computable numbers list).
0.1234 98765...
and the sequence in the suffix is NOT anywhere on CNL?
Barb says yes, Will says no, you're playing non commital for as long as you can, as always.
Herc
0.123498765 is computable, but there is a number 0.123498765.... that is
not computable.
The possible answers are: "no, 1 isn't a member of the set so it isn't
plotted" or "the question doesn't make sense, what do you mean by
'plot'?" or "1 is the limit of the points plotted, if that's what you
mean then 'yes'".
is the point 1.0 plotted?
Herc
no. 1 is not in the set {.9, .99, .999, ...}, so it is not plotted.
You keep asking this question and keep getting the same answer. Do you
really expect it to change?
All prefixes are on the CNL. Any member of T_b:
T_b = {k / b^n, k, b, n in J, b > 1, n > 0, 0 <= k < b^n}
is on the CNL -- the obvious TM would have (n+1) states and
a transition function that writes out the appropriate digit
and shifts to the next state.
Any member of Q intersect [0,1] is on the CNL, as one can
construct a TM to spit out digits; the operation of long
division is in effect a moving TM, with the remainder
being the state. Pick a base.
Algebraic numbers within [0,1] are probably on the CNL,
though I'd have to do some work to construct an obvious
TM; the usual formulation is to use numeric approximation
with an initial guess. Certainly all algebraic numbers
are computable using other means (e.g., infinite integer
machine language systems).
Is an arbitrary number within [0,1] on the CNL?
It's not a guarantee as the number of states of the obvious TM
will become infinite, which means it's no longer a TM, but
an oracle. However, I'm not sure there's an existence proof
beyond the ones already shown, such as both of Cantor's proofs,
which you've previously rejected.
Your confusion will continue. :-) (One needn't be psychic to
predict that.)
>
> Herc
So what is the number ("+" + 9) / 10?
How about "+" * 10 - 9?
>
> You're welcome Herc.
>
> Garry
that's not confusion, that's interpreting your posts in relation to the question.
Herc