Hamel Basis of R over Q
Maury Barbato
consider R as a vector space over the field Q.
R cannot have a finite basis, otherwise it would be
countable. Moreover, R cannot have a countable Hamel
basis, otherwise it would be a countable union of
countable sets (the subspaces spanned by the first
n elements of the basis), and so it would be countable
again.
My question is: can we prove that a Hamel basis of R
has cardinality c(the continuum)?
This is a purely set-theoretic question, and I have
no idea of the axiomatic set theory.
Thank you very much for your attention.
My Best Regards,
Maury Barbato
master1729
i intended too post a similar text today.
bravo ! you have found the disproof of AC.
AC is indeed inconsistant with cardinality.
regards
the master
tommy1729
David Hartley
<340140750.43720.12578...@gallium.mathforum.org>,
Maury Barbato <maurizi...@aruba.it> writes
>My question is: can we prove that a Hamel basis of R has cardinality
>c(the continuum)? This is a purely set-theoretic question, and I have
>no idea of the axiomatic set theory.
Yes.
If the basis has cardinality k, then it has k finite subsets. These each
generate a countable subspace, and the union of these subspaces is all
of R. So |R| <= k*aleph_0 = k <= |R|.
(Assuming AC of course; without it even your proof that the basis is
uncountable doesn't hold.)
--
David Hartley
David C. Ullrich
<maurizi...@aruba.it> wrote:
>Hello,
>consider R as a vector space over the field Q.
>R cannot have a finite basis, otherwise it would be
>countable. Moreover, R cannot have a countable Hamel
>basis, otherwise it would be a countable union of
>countable sets (the subspaces spanned by the first
>n elements of the basis), and so it would be countable
>again.
>
>My question is: can we prove that a Hamel basis of R
>has cardinality c(the continuum)?
Yes. Say the cardinality of the basis is alpha.
Since alpha^2 = alpha for any infinite cardinal
it follows that the cardinality of R is
alpha + alpha^2 + alpha^3 + ...
= alpha + alpha + ...
= alpha.
>This is a purely set-theoretic question, and I have
>no idea of the axiomatic set theory.
>
>Thank you very much for your attention.
>My Best Regards,
>Maury Barbato
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
David C. Ullrich
wrote:
Why do you babble nonsense like this?
>
>regards
>
>the master
>
>tommy1729
master1729
its the truth.
AC is simply false.
if you dont see the relationship of the OP with AC then you really need to open a book more often.
thus its far from nonsense , its mathematically related and important.
>
> >
> >regards
> >
> >the master
> >
> >tommy1729
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
tommy1729
master1729
> On Tue, 10 Nov 2009 05:53:37 EST, Maury Barbato
> <maurizi...@aruba.it> wrote:
>
> >Hello,
> >consider R as a vector space over the field Q.
> >R cannot have a finite basis, otherwise it would be
> >countable. Moreover, R cannot have a countable Hamel
> >basis, otherwise it would be a countable union of
> >countable sets (the subspaces spanned by the first
> >n elements of the basis), and so it would be
> countable
> >again.
> >
> >My question is: can we prove that a Hamel basis of R
>
> >has cardinality c(the continuum)?
>
> Yes. Say the cardinality of the basis is alpha.
>
> Since alpha^2 = alpha for any infinite cardinal
> it follows that the cardinality of R is
>
> alpha + alpha^2 + alpha^3 + ...
>
> = alpha + alpha + ...
>
> = alpha.
i wonder were david got this from.
a book ?
despite considering AC nonrelated and nonsensical compared to the OP ( see his reply to me ), he does give a correct proof.
thats a huge paradox.
either he got the proof from somewhere or he knows his stuff and is thus deliberately lying about the relation with AC.
>
> >This is a purely set-theoretic question, and I have
> >no idea of the axiomatic set theory.
> >
> >Thank you very much for your attention.
> >My Best Regards,
> >Maury Barbato
>
> David C. Ullrich
>
> "Understanding Godel isn't about following his formal
> proof.
> That would make a mockery of everything Godel was up
> to."
> (John Jones, "My talk about Godel to the post-grads."
> in sci.logic.)
tommy1729
Maury Barbato
Thank you so much for your help, David!
Friendly Regards,
Maury Barbato
Maury Barbato
> In message
> <340140750.43720.1257850447624.JavaMail.root@gallium.m
Thank you very very much!
David C. Ullrich
wrote:
>david wrote :
>
>> On Tue, 10 Nov 2009 05:53:37 EST, Maury Barbato
>> <maurizi...@aruba.it> wrote:
>>
>> >Hello,
>> >consider R as a vector space over the field Q.
>> >R cannot have a finite basis, otherwise it would be
>> >countable. Moreover, R cannot have a countable Hamel
>> >basis, otherwise it would be a countable union of
>> >countable sets (the subspaces spanned by the first
>> >n elements of the basis), and so it would be
>> countable
>> >again.
>> >
>> >My question is: can we prove that a Hamel basis of R
>>
>> >has cardinality c(the continuum)?
>>
>> Yes. Say the cardinality of the basis is alpha.
>>
>> Since alpha^2 = alpha for any infinite cardinal
>> it follows that the cardinality of R is
>>
>> alpha + alpha^2 + alpha^3 + ...
>>
>> = alpha + alpha + ...
>>
>> = alpha.
>
>i wonder were david got this from.
>
>a book ?
Huh? Most math that most people know they learned from books.
>despite considering AC nonrelated
Huh? Where did I say that AC was unrelated to this question?
>and nonsensical
Huh? You're saying that I consider AC nonsensical? Nonsense.
Your _reply_ to the OP was babbling nonsense, as is this post.
William Hughes
> Hello,
> consider R as a vector space over the field Q.
> R cannot have a finite basis, otherwise it would be
> countable. Moreover, R cannot have a countable Hamel
> basis, otherwise it would be a countable union of
> countable sets (the subspaces spanned by the first
> n elements of the basis), and so it would be countable
> again.
>
> My question is: can we prove that a Hamel basis of R
> has cardinality c(the continuum)?
As far as I can see you just did
A Hamel basis cannot be countable.
A Hamel basis is a subset of the reals
so it must have cardinality less
than or equal to c.
Am I missing something?
- William Hughes
A
Isn't the Continuum Hypothesis precisely the statement that there
exists no set of cardinality strictly between that of the integers and
that of the reals? It seems like that is what is missing from your
argument--one needs to know that the cardinality of the Hamel basis
isn't some uncountable cardinal less than that of the real numbers.
David Hartley
<55b415c3-1ec0-421f...@g23g2000yqh.googlegroups.com>,
William Hughes <wpih...@hotmail.com> writes
>> My question is: can we prove that a Hamel basis of R
>> has cardinality c(the continuum)?
>
>
>
>As far as I can see you just did
>
>A Hamel basis cannot be countable.
>
>A Hamel basis is a subset of the reals
>so it must have cardinality less
>than or equal to c.
>
>Am I missing something?
Yes, cardinals between aleph_0 and c.
--
David Hartley
master1729
> lies lies lies
anyone who can read my posts , the op and his posts in this tread can see the contradiction.
David C. Ullrich
wrote:
>david wrote
>
>> lies lies lies
>
>anyone who can read my posts , the op and his posts in this tread can see the contradiction.
You realize of course that you're saying that every mathematician on
te planet is simply wrong. Not only that, but it's not even like
they're wrong about something tricky, anyone who can read can
see that every mathematician on the planet is wrong.
And you wonder why you don't get more respect.
Ki Song
> On Wed, 11 Nov 2009 10:14:46 EST, master1729 <tommy1...@gmail.com>
Hi David,
With all due respect, why do you bother responding to this imbecile?
William Hughes
You realize that by editing a quote.
you run the risk of losing respect.
Right, no down side.
- William Hughes
David C. Ullrich
You seem to be assuming the Coninuum Hypothesis...
> - William Hughes
David C. Ullrich
Everyone needs a hobby.
William Hughes
> On Wed, 11 Nov 2009 05:26:01 -0800 (PST), William Hughes
>
>
>
> <wpihug...@hotmail.com> wrote:
> >On Nov 10, 6:53 am, Maury Barbato <mauriziobarb...@aruba.it> wrote:
> >> Hello,
> >> consider R as a vector space over the field Q.
> >> R cannot have a finite basis, otherwise it would be
> >> countable. Moreover, R cannot have a countable Hamel
> >> basis, otherwise it would be a countable union of
> >> countable sets (the subspaces spanned by the first
> >> n elements of the basis), and so it would be countable
> >> again.
>
> >> My question is: can we prove that a Hamel basis of R
> >> has cardinality c(the continuum)?
>
> >As far as I can see you just did
>
> >A Hamel basis cannot be countable.
>
> >A Hamel basis is a subset of the reals
> >so it must have cardinality less
> >than or equal to c.
>
> >Am I missing something?
>
> You seem to be assuming the Coninuum Hypothesis...
Yeah, kind of silly in context.
Mea Culpa
- William Hughes
master1729
ok. you have a point there.
but not all mathematicians are wrong.
since not all mathematicians are in favor of AC.
what is obvious is the ' relationship to AC ' , surely you agree on that ?
i will need to explain further why AC is wrong , i know.
but that is mainly for clarity for those who dont understand the arguments against AC very well.
master1729
> In message
> <55b415c3-1ec0-421f...@g23g2000yqh.goog
> legroups.com>,
> William Hughes <wpih...@hotmail.com> writes
> >> My question is: can we prove that a Hamel basis of
> R
> >> has cardinality c(the continuum)?
> >
> >
> >
> >As far as I can see you just did
> >
> >A Hamel basis cannot be countable.
> >
> >A Hamel basis is a subset of the reals
> >so it must have cardinality less
> >than or equal to c.
> >
> >Am I missing something?
>
>
> Yes, cardinals between aleph_0 and c.
> --
> David Hartley
huh ??
AC and CH are usually considered different.
at least in standard set theories.
in fact , i think i alone ever considered AC = ~ CH.
and as far as i know , nobody believes in AC = CH.
(CH = continuum hypothesis in case you dont know )
if you have a personal nonstandard theory im intrested though , could overlap with my own ideas though.
regards
tommy1729
master1729
your an imbicile ki song , with only 4 posts containing nothing meaningfull.
at least david is a worthy opponent.
you with your 4 posts are nothing !
David C. Ullrich
wrote:
>david wrote :
>
>> On Wed, 11 Nov 2009 10:14:46 EST, master1729
>> <tomm...@gmail.com>
>> wrote:
>>
>> >david wrote
>> >
>> >> lies lies lies
>> >
>> >anyone who can read my posts , the op and his posts
>> in this tread can see the contradiction.
>>
>> You realize of course that you're saying that every
>> mathematician on
>> te planet is simply wrong. Not only that, but it's
>> not even like
>> they're wrong about something tricky, anyone who can
>> read can
>> see that every mathematician on the planet is wrong.
>>
>> And you wonder why you don't get more respect.
>>
>>
>>
>>
>
>ok. you have a point there.
>
>but not all mathematicians are wrong.
>
>since not all mathematicians are in favor of AC.
Not that this has any bearing on the silly things
you've said here, but give an example of a
mathematician who's not "in favor of" AC.
>what is obvious is the ' relationship to AC ' , surely you agree on that ?
Asking whether I agree that the relationship to AC is obvious
is much too vague.
What started this was your assertion "AC is indeed inconsistant
with cardinality." That's nonsense.
It's nonsense at least twice. It's nonsense because nothing
in this thread gives any evidence of any such thing.
And it's nonsense that shows your ignorance of
mathematics: In fact AC is _needed_ for the standard
_definition_ of cardinality!
>i will need to explain further why AC is wrong , i know.
>
>but that is mainly for clarity for those who dont understand the arguments against AC very well.
Right. You should also explain further why the Earth is flat, for
whose who don't understand the arguments against the round
Earth theory very well.
David Hartley
<2096606398.8012.12580...@gallium.mathforum.org>,
master1729 <tomm...@gmail.com> writes
>david hartley wrote :
>
>> In message
>> <55b415c3-1ec0-421f...@g23g2000yqh.goog
>> legroups.com>,
>> William Hughes <wpih...@hotmail.com> writes
>> >> My question is: can we prove that a Hamel basis of
>> R
>> >> has cardinality c(the continuum)?
>> >
>> >
>> >
>> >As far as I can see you just did
>> >
>> >A Hamel basis cannot be countable.
>> >
>> >A Hamel basis is a subset of the reals
>> >so it must have cardinality less
>> >than or equal to c.
>> >
>> >Am I missing something?
>>
>>
>> Yes, cardinals between aleph_0 and c.
>huh ??
>
>AC and CH are usually considered different.
>
>at least in standard set theories.
>
>in fact , i think i alone ever considered AC = ~ CH.
>
>and as far as i know , nobody believes in AC = CH.
>
>(CH = continuum hypothesis in case you dont know )
>
>if you have a personal nonstandard theory im intrested though , could
>overlap with my own ideas though.
I was merely reminding William that CH was not assumed in the original
question. How on earth can you read that to imply AC and CH are
equivalent?
--
David Hartley
master1729
i dont need AC to define cardinality.
I use the Powerset.
as in ' the powerset of N is R '.
why involve AC ?
master1729
> <2096606398.8012.1258069173297.JavaMail.root@gallium.m
that doesnt make sense.
CH was not mentioned by William.
you mentioned CH.
A
> > On Thu, 12 Nov 2009 18:30:57 EST, master1729
> > wrote:
>
> > >david wrote :
>
> > >> On Wed, 11 Nov 2009 10:14:46 EST, master1729
How do you define cardinality using just the power set?
David Hartley
<1960616631.35165.12583...@gallium.mathforum.org>,
master1729 <tomm...@gmail.com> writes
>that doesnt make sense.
>
>CH was not mentioned by William.
>
>you mentioned CH.
William asked what was wrong with his simple answer to the original
question. I, and others, pointed out that it assumed there were no
cardinals strictly between aleph_0 and c, i.e. that he was implicitly
assuming the CH, (although I didn't actually mention it by name).
--
David Hartley
Jim Burns
> david wrote :
[...]
>> What started this was your assertion "AC is indeed
>> inconsistant
>> with cardinality." That's nonsense.
>>
>> It's nonsense at least twice. It's nonsense because
>> nothing
>> in this thread gives any evidence of any such thing.
>> And it's nonsense that shows your ignorance of
>> mathematics: In fact AC is _needed_ for the standard
>> _definition_ of cardinality!
[...]
> i dont need AC to define cardinality.
>
> I use the Powerset.
>
> as in ' the powerset of N is R '.
>
> why involve AC ?
AC is need to show that, for all sets A and B,
card(A) =< card(B) and/or card(A) >= card(B).
Jim Burns
Herman Rubin
A <anonymous....@yahoo.com> wrote:
>On Nov 15, 5:28=A0pm, master1729 <tommy1...@gmail.com> wrote:
>> > On Thu, 12 Nov 2009 18:30:57 EST, master1729
>> > <tommy1...@gmail.com>
>> > wrote:
>> > >david wrote :
>> > >> On Wed, 11 Nov 2009 10:14:46 EST, master1729
>> > >> <tommy1...@gmail.com>
>> > >> wrote:
>> > >> >david wrote
................
>> i dont need AC to define cardinality.
>> I use the Powerset.
>> as in ' the powerset of N is R '.
>> why involve AC ?
>How do you define cardinality using just the power set?
Defining cardinality is not of much, if any, use.
One can define ordering and equality of cardinal
numbers, and that is enough; just assume there is
such an animal and use it. It will not lead to
problems; how do you think that theorems on cardinal
numbers related to the axiom of choice are proved?
This is from someone who has proved many.
--
This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hru...@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
A
> In article <1fe3d1df-5b0f-4533-82f6-9a1f87452...@e31g2000vbm.googlegroups.com>,
>
> A <anonymous.rubbert...@yahoo.com> wrote:
> >On Nov 15, 5:28=A0pm, master1729 <tommy1...@gmail.com> wrote:
> >> > On Thu, 12 Nov 2009 18:30:57 EST, master1729
> >> > <tommy1...@gmail.com>
> >> > wrote:
> >> > >david wrote :
> >> > >> On Wed, 11 Nov 2009 10:14:46 EST, master1729
> >> > >> <tommy1...@gmail.com>
> >> > >> wrote:
> >> > >> >david wrote
>
> ................
>
> >> i dont need AC to define cardinality.
> >> I use the Powerset.
> >> as in ' the powerset of N is R '.
> >> why involve AC ?
> >How do you define cardinality using just the power set?
>
> Defining cardinality is not of much, if any, use.
>
> One can define ordering and equality of cardinal
> numbers, and that is enough; just assume there is
> such an animal and use it. It will not lead to
> problems; how do you think that theorems on cardinal
> numbers related to the axiom of choice are proved?
>
> This is from someone who has proved many.
I wasn't expecting to learn something about set theory from a reply
from "master1729"--I was just hoping to get him to make more of his
entertaining posts.
David C. Ullrich
wrote:
Huh? Saying the powerset of N is R is not a definition
of cardinality.
A definition of cardinality looks like this:
If S is a set then card(S) = ... .
Exactly how do you finish that definition?
>why involve AC ?
David C. Ullrich
Rubin) wrote:
>In article <1fe3d1df-5b0f-4533...@e31g2000vbm.googlegroups.com>,
>A <anonymous....@yahoo.com> wrote:
>>On Nov 15, 5:28=A0pm, master1729 <tommy1...@gmail.com> wrote:
>>> > On Thu, 12 Nov 2009 18:30:57 EST, master1729
>>> > <tommy1...@gmail.com>
>>> > wrote:
>
>>> > >david wrote :
>
>>> > >> On Wed, 11 Nov 2009 10:14:46 EST, master1729
>>> > >> <tommy1...@gmail.com>
>>> > >> wrote:
>
>>> > >> >david wrote
>
> ................
>
>>> i dont need AC to define cardinality.
>
>>> I use the Powerset.
>
>>> as in ' the powerset of N is R '.
>
>>> why involve AC ?
>
>
>>How do you define cardinality using just the power set?
>
>Defining cardinality is not of much, if any, use.
>
>One can define ordering and equality of cardinal
>numbers, and that is enough;
One can certainly _define_ ordering of cardinals
without defining cardinals themselves, or by defining
them in some way without AC (for instance as
proper classes).
But how does one show that any two
carinalities are comparable without AC?
>just assume there is
>such an animal and use it. It will not lead to
>problems; how do you think that theorems on cardinal
>numbers related to the axiom of choice are proved?
>
>This is from someone who has proved many.
David C. Ullrich
David Hartley
Ullrich <dull...@sprynet.com> writes
>One can certainly _define_ ordering of cardinals without defining
>cardinals themselves, or by defining them in some way without AC (for
>instance as proper classes).
>
>But how does one show that any two
>carinalities are comparable without AC?
You don't.
That is, "comparability of cardinalities" is equivalent to AC.
(Given a set, you can always form an ordinal which is not smaller,
without using choice. By "comparability" the ordinal must be bigger,
i.e. there is an injection from the set into the ordinal, and so the set
can be well-ordered.)
--
David Hartley
David C. Ullrich
wrote:
Thanks - actually it was a rhetorical question.
master1729
wrong , statements like =< or >= are trivial since
<,>,= are the only possibilities.
Herman Rubin is correct.
master1729
Wrong !
Herman Rubin is correct.
wrong because one can easily compare countable and uncountable sets , prove that they are different ( e.g. cantors diagonal ) and still not use AC !!
master1729
but you rubbertube are the one who is entertaining , I agree with Herman Rubin , whereas you stand corrected by Herman Rubin , in an attempt to correct me.
very funny.
A
I wasn't trying to correct you! And Herman Rubin did not answer the
question I asked you: how do you define cardinality using only the
power set?
What I am really curious about here is what it looks like when you,
you personally, actually write down some definitions and maybe even
some proofs. This thread (and others) are full of people trying to
correct you, or telling you that you are wrong about various things--
of course, if you begin from definitions and prove the statements you
make, that ends most arguments immediately. But you don't seem to be
trying to do this. I was hoping to get you to write down a definition
and a proof--in this case, a definition of cardinality or cardinal
number, and a proof that such a thing is well-defined and has
appropriate basic properties coming from the power set operation--but
actually I would be interested in seeing you write down a precise
definition and a rigorous proof of anything else related to your ideas
about set theory.
For example, in a different, very recent post in this thread, when
somebody says that the axiom of choice is equivalent to the
comparability of any two cardinal numbers, you say this:
> Wrong !
>
> Herman Rubin is correct.
>
> wrong because one can easily compare countable and uncountable sets , prove that they are different ( e.g.
> cantors diagonal ) and still not use AC !!
I do not think Herman Rubin made any such claim. But let's get precise
here: in saying that two sets are "comparable" we mean that the
cardinality of one set is either greater than, equal to, or less than
the cardinality of the other set; this requires that we be able to
inject one set into the other. Since you object to the notion that the
axiom of choice is required to have comparability of sets, suppose X
and Y are two sets. How do you propose to show that one has
cardinality greater than or less than the other, i.e., how do you
propose to show that there exists an injection from one to the other,
without the axiom of choice?
master1729
yes you were.
And Herman Rubin did
> not answer the
> question I asked you: how do you define cardinality
> using only the
> power set?
herman confirmed AC was not needed.
this means there are many ways to define card without AC.
you are not curious.
your just waiting for me to post something you can disagree on or laugh at.
i did not mention injection , and it is not needed.
one can show uncountable > countable by using for instance cantors diagonal argument ( base 3 ).
see ! no AC !
tommy1729
the master
A
Yes, there are many ways to define cardinality without the axiom of
choice. You said you use the power set to define cardinality. Does
this mean you use the power set operation to put a partial ordering on
the cardinal numbers? Or does it mean you use the power set operation
to define cardinal numbers themselves? I am still hoping you will
explain what you meant.
I'm not asking what other people do, or what things are standard,
because I'm pretty sure I already know what's standard. You seem to be
claiming to do something new and interesting, but you haven't given
details yet--I'm asking what YOU mean when you say you define
cardinality using the power set.
I don't think I'm as mean-spirited as you think I am. I think your
posts are entertaining, because you make grandiose claims (like
claiming you have proven the Riemann Hypothesis) with a lot of
showmanship, calling yourself "the master" and talking down to
everyone else, but I think you're just as aware of that as I am, and
it seems like you're just as entertained by it as I am--it's fun to
read, and I imagine it's fun to write. But you backpedal or rely on
others when you're asked for details about these claims--you'd be a
lot more convincing if you could supply some rigorous definitions and
proofs.
Something like this is true--you can compare sufficiently small sets
with weakened forms of the axiom of choice, or even without the axiom
of choice at all (e.g. finite sets can be compared in ZF). But your
statement above, which I think is supposed to mean that the
cardinality of any countable set is strictly less than the cardinality
of any uncountable set: tell me how that works. Suppose X is a
countable set (it admits a bijection with the natural numbers) and
suppose Y is an uncountable set (it admits no injection into the
natural numbers). How do we know that there exists an injection from X
into Y, without any form of the axiom of choice?
A
Wait a minute, now I am re-reading this: you say that injection is
"not needed." Let X and Y be sets. What do you mean when you say that
X has cardinality greater than or equal to that of Y? Usually this
means precisely that there exists an injection from Y to X. How are
you ordering the cardinal numbers without using injections?
James Burns
> Jim Burns wrote :
>>master1729 wrote:
>>>david wrote :
>>>>What started this was your assertion "AC is indeed
>>>>inconsistant with cardinality." That's nonsense.
>>>>
>>>>It's nonsense at least twice. It's nonsense
>>>>because nothing in this thread gives any
>>>>evidence of any such thing.
>>>>
>>>>And it's nonsense that shows your ignorance of
>>>>mathematics: In fact AC is _needed_ for the
>>>>standard _definition_ of cardinality!
[...]
>>>why involve AC ?
>>
>>AC is need to show that, for all sets A and B,
>>card(A) =< card(B) and/or card(A) >= card(B).
> wrong , statements like =< or >= are trivial
> since <,>,= are the only possibilities.
Have you tried proving that without using AC?
Jim Burns
David C. Ullrich
wrote:
You started this by saying that "AC is indeed inconsistant with
cardinality."
That's nonsense.
I said AC was needed to define cardinality. It is indeed needed
for the stanadard definition of card(S). Herman pointed out
that one doesn't need that particular definition, fine.
Regardless of the definition, AC _is_ needed to show that
any two cardinalities are comparable. Which is sort of
an important aspect of cardinality. And which shows
that your claim that AC is "inconcistsnt with" cardinality
is just silly.
Then you said you define cardinality using power set.
Several people have asked how you do that. You
haven't replied.
Sorry. You seem to be under the impression that when
Herman corrects something I said that proves that you
were making sense. That's not so. The only point to
Hermann's correction is that I should have said
"the standard definition of card(S)" instead of
"cardinality" - of course that's what I meant.
That doesn't change the fact that your statement
that AC "contradicts" cardinality was stupid.
And it doesn't change the fact that you haven't
given any explanation of what you meant when
you said you "use powerset" to define cardinality.
David C. Ullrich
wrote:
Erm, show us how to prove that.
First you have to give us your _definitions_ of
"card(S) < card(T)", "card(S)=card(T)"
and "card(S) > card(T)". Then prove that
for any two sets S and T one of those holds.
Good luck, by the way - the fact that one of those
must hold _is_ equivalent to AC.
>
>Herman Rubin is correct.
Everything he said was correct, yes.
Nothing he said shows that anything you've
said here is correct. That includes three
statments of yours: AC contradicts cardinality,
you use powerset to define cardinality, and
<, > and = are the only possibilities.
master1729
see my other replies.
master1729
a = b
a > b
a < b
its trivial. there are no other possibilities. prehistoric math.
master1729
in order of size
R , 2^R , 2^2^R , 2^2^2^R , ...
ive said that many times before btw.
like it or not , but the above makes perfect sense.
no AC needed. no ordinals needed. no CH needed.
master1729
see my other replies of today.
tommy1729
Virgil
<a81560c4-87fb-4b7f...@l2g2000yqd.googlegroups.com>,
Ki Song <kiwis...@gmail.com> wrote:
> What are a and b? Are they positive integers? If that's the case,
> then you have no idea what you are talking about.
If a and b are merely sets, with "=" meaning "having the same members
and "<" and ">" meaning proper subset and proper superset respectively,
then there are certainly instances in which none of a = b, a > b or
a < b hold.
More commonly, however, for sets a and b, a < b is taken to mean that
there is an injective mapping from set a to set b, but none from b to a.
and a = b is taken to mean that there is a bijection between a and B.
And in this latter case, the axiom of choice is needed to establish
trichotomy for "<", ">" and "=".
A
Here is another possibility: a is not equal to b, a is not greater
than b, and a is not less than b. The ordering on the cardinal numbers
is a partial ordering, but it is by no means clear that, given any two
sets A and B, the cardinal number of A is less than, equal to, or
greater than the cardinal number of B; perhaps the cardinal number of
A is incomparable with the cardinal number of B.
How about this: you claim that it's trivial, and "prehistoric," that
given any two cardinal numbers, either they are equal to one another
or one is greater than the other. I agree that any two cardinal
numbers are comparable if one assumes the axiom of choice, but I want
to know how you propose to show that it is true without using the
axiom of choice. I would be very happy if you would do one of the
following things: either provide a proof that any two cardinal numbers
are comparable without using the axiom of choice, or give a reference
of somewhere (a paper or a book) where this has already been proven,
or even stated.
A
Let X be a set. How do we know that there exists a bijection between X
and some set of the form 2^2^2^...^2^R? I agree that you have put a
very reasonable partial ordering on all sets of the form
2^2^2^...^2^R, but it is by no means clear that every set is of that
form.
David C. Ullrich
wrote:
>cardinalities larger or equal to R by powersets :
>
>in order of size
>
>R , 2^R , 2^2^R , 2^2^2^R , ...
>
>
>ive said that many times before btw.
That's supposed to be a definition of cardinality?
>like it or not , but the above makes perfect sense.
It doesn't answer the question: Given a set S,
what is card(S)?
>no AC needed. no ordinals needed. no CH needed.
This is funnier than usual. If we try to make sense of your
"definition" of cardinality the best we can do is this:
Given an infinite set S, find the set in that sequence
above that has the same size as S (in terms of the
existence of a bijection). But first, it's a fact that there
are sets that are "larger" than any of the sets in that
sequence, and second (this is the really funny part)
saying that there is no set "in between" R and 2^R
is _precisely_ what CH says!
David C. Ullrich
wrote:
If it's trivial you should be able to give a proof.
(It _cannot_ be proved without AC.)
Aatu Koskensilta
> no AC needed. no ordinals needed. no CH needed.
How do you propose to show that every uncountable cardinal is a Beth
number?
--
Aatu Koskensilta (aatu.kos...@uta.fi)
"Wovon man nicht sprechen kann, dar�ber muss man schweigen"
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
Aatu Koskensilta
> see my other replies of today.
Why? They consist solely of nonsensical one-liners.