Good evening,

I want to show that all bases of a vector space have the same cardinality, and it needs the following equality : Let $\aleph_0$ be the cardinality of $\mathbb{N}$ and $X$ an infinite set, then $$\aleph_0 \operatorname{card}(X) = \operatorname{card}(X).$$

Does anyone know where there is a proof for this equality?

Thanks in advance,

Duc Anh

  • $\begingroup$ Thank you, i corrected it. $\endgroup$ – Đức Anh Sep 5 '12 at 12:44
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    $\begingroup$ I'm not sure how low-level you expect the proof to be. Assuming axiom of choice, any infinite cardinality is at least $\aleph_0$, and for any two cardinals $\kappa,\lambda$, $\kappa\cdot\lambda=\max(\kappa,\lambda)$. $\endgroup$ – tomasz Sep 5 '12 at 12:46
  • $\begingroup$ Thank you, and how to prove your equality? I know the Axiom of choice. $\endgroup$ – Đức Anh Sep 5 '12 at 12:48
  • $\begingroup$ @tomasz: for any two infinite cardinals. $\endgroup$ – Henning Makholm Sep 5 '12 at 12:49
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    $\begingroup$ @HenningMakholm: any two cardinals, at least one of which is infinite, if we're being picky. ;) $\endgroup$ – tomasz Sep 5 '12 at 12:59

We use the fact that $|X\times X|=|X|$, applying Zorn lemma to $S:=\{(B,f), B\subset X, f\colon B\times B\to B,f\mbox{ bijective}\}$ with the partial order $(B_1,f_1)\leq (B_2,f_2)$ if $B_1\subset B_2$ and $f_{2\mid B_1}=f_1$. Then we conclude by Cantor-Bernstein theorem.

  • $\begingroup$ Thanks, i'm trying to understand your proof. I will ask you later :D $\endgroup$ – Đức Anh Sep 5 '12 at 13:08
  • $\begingroup$ @tomasz Thanks, I have edited. $\endgroup$ – Davide Giraudo Sep 5 '12 at 13:18
  • $\begingroup$ This assumes we that already know that $|2\times X|=|X|$, right? Or how do you conclude that a maximal element has $B=X$? $\endgroup$ – Henning Makholm Sep 5 '12 at 13:29
  • $\begingroup$ @HenningMakholm Yes, I think we need this intermediate step (which also uses Zorn lemma if I'm not wrong). $\endgroup$ – Davide Giraudo Sep 5 '12 at 13:31
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    $\begingroup$ Thank you for the answer, I have found a proof in Hewitt, Stromberg, Real and Abstract Analysis. $\endgroup$ – Đức Anh Sep 5 '12 at 20:08

Here's a more direct approach; I think something like this must be hidden in the intermediate step of Davide's proof anyway:

First, we need to assume the Axiom of Choice. So $X$ can be well-ordered, and without loss of generality we can assume that $X$ is an initial ordinal. In particular, then, $X$ is a limit ordinal.

One can easily prove that every ordinal can be uniquely written as $\alpha+n$ where $\alpha$ is zero or a limit ordinal and $n$ is finite.

Fix a bijection $f:\mathbb N\times\mathbb N\to \mathbb N$. Then $$ \langle n,\alpha+m\rangle \mapsto \alpha+f(n,m) $$ is a bijection $\mathbb N\times X \to X$.

  • $\begingroup$ Thank you, it seems to me that your proof will work well, but I have to learn the notion of ordinal (so this is a very good chance to do it). I will ask you later. $\endgroup$ – Đức Anh Sep 5 '12 at 18:38
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    $\begingroup$ If you want to do anything serious in set theory -- beyond memorizing some facts about cardinality and knowing how to formalize other mathematics within set theory as a background framework -- then you will need to learn about ordinals, yes. They are the scaffolding everything else hangs from. $\endgroup$ – Henning Makholm Sep 5 '12 at 19:12
  • $\begingroup$ Thank you for the advice. I have read the notion of ordinal numbers. But I don't understand your formula : what is $\alpha + m$ and how can $\alpha + m $ be considered as an element of $X$ ? $\endgroup$ – Đức Anh Sep 5 '12 at 20:20
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    $\begingroup$ $\alpha+m$ is the ordinal sum of $\alpha$ and $m$. An ordinal is the set of all ordinals smaller than it. Since I'm assuming that $X$ is an ordinal, its members are other, smaller, ordinals, and so they can be written as $\alpha+m$. $\endgroup$ – Henning Makholm Sep 5 '12 at 20:22
  • $\begingroup$ Ah, I think I have understood your solution. Thank you very much. $\endgroup$ – Đức Anh Sep 5 '12 at 20:27

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