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Let $V$ be a vector space over a field $\mathbb{K}$, such that $V$ admits an infinite linearly independent set. Let $B$ and $B'$ be two bases for $V$. Then $|B|=|B'|$.

My approach:

Let $\mathcal{F}(B)$ and $\mathcal{F}(B')$ be the collections of all finite subsets of $B$ and $B'$. The idea is to conclude that $|\mathcal{F}(B)|=|\mathcal{F}(B')|$, from which we can deduce that $|B|=|B'|$ (since $|\mathcal{F}(B)|=|B|$).

There is a theorem which says that if there is a countable-to-one function $\phi: X\to Y$ between two sets $X$ and $Y$ then $|X|\le \aleph_0 |Y|$.

To this end, suppose, without loss of generality, that $|\mathcal{F}(B)|\le \mathcal{F}(B')$. Then there is a surjective function $\phi$ from $\mathcal{F}(B')$ to $\mathcal{F}(B)$. Thus for $x\in \mathcal{F}(B)$, $\phi^{-1}(\{x\})$ exists and is countable (since $\mathcal{F}(B')$ is countable). Hence, $|\mathcal{F}(B')|\le \aleph_0 |\mathcal{F}(B)|=|\mathcal{F}(B)|$. So that $|B|=|\mathcal{F}(B)|=|\mathcal{F}(B')|=|B'|$, as required.

Please let me know whether my proof is fine. I'm not very confident in this field yet.

Asaf Karagila
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sequence
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  • Why is $\mathcal{F}(B')$ countable? – Mr. Chip Sep 21 '17 at 23:15
  • Because this is the set of all finite subsets of $B'$. But a finite subset contains finitely many elements. But we can count them by their finite cardinality. Now I'm not so certain, however, if it's countable or not. Would appreciate some hints. – sequence Sep 22 '17 at 00:02
  • Yeah but being infinite can mean being a lot bigger than countable... what if $|B| = \mathbb{R}$? – Mr. Chip Sep 22 '17 at 00:07
  • My intuition was that, since all elements of $\mathcal{F}(B')$ are countable sets themselves, $\mathcal{F}(B')$ itself is countable. – sequence Sep 22 '17 at 00:12
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    I mean, suppose $B$ has size at least that of $\mathbb{R}$. Then $\mathcal{F}(B)$ contains, in particular, the singleton ${b}$ for every $b \in B$, so... – Mr. Chip Sep 22 '17 at 00:13
  • Good point. Would you give me a suggestion on how to revise this proof? – sequence Sep 22 '17 at 00:17

1 Answers1

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First prove the following lemma.

Let $A$ be a set, $\eta$ be a cardinal number, and $(B_a)_{a\in A}$ be a family of sets satisfying $|B_a|\le\eta$ for all $a\in A$. Then $$ \left| \cup_{a\in A}B_a \right| \le |A|\eta. $$

Sketch of proof in spoiler below:

Without loss of generality, suppose $(B_a)_{a\in A}$ are pairwise disjoint. Choose an injection $g_a:B_a\to E$ for each $a\in A$, where $E$ is set of cardinality $\eta$. Define an injection from $\cup_{a\in A}B_a\to A\times E$ by mapping $x$ in the union to the element $(a,g_a(x))$ in $A\times E$, where $a\in A$ is such that $x\in B_a$. Prove this is a well-defined injection.

Now suppose $A$ and $B$ are infinite bases of a vector space. For each $a\in A$ find a finite subset $B_a$ of $B$ such that $a$ is in the span of $B_a$. Then $B=\cup_{a\in A}B_a$. Use this and the lemma to obtain $|B|\le|A|$. More details are in the spoiler below:

We obtain $B=\cup_{a\in A}B_a$ since $B$ is linearly independent and $A$ is spanning. Using the lemma, we have $$ |B|=|\cup_{a\in A}B_a|\le|A|\aleph_0=|A|, $$ because $|B_a|<\aleph_0$ for each $a\in A$. Then repeat the same argument to get $|A|\le|B|$, and conclude using the Cantor-Schröder-Bernstein theorem.

John Griffin
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  • If $A$ is uncountable, could we still say that $|B_a|\le |\mathbb{N}|$? – sequence Sep 22 '17 at 01:19
  • @sequence Yes. The reason $|B_a|\le|\mathbb{N}|$ for each $a\in A$ is because we chose $B_a$ to be finite. We can always do this because in order for $B$ to be a basis, every element in the vector space must be in the span of a finite subset of $B$. – John Griffin Sep 22 '17 at 01:22
  • Is it true that $\mathcal{F}(A)$ is countable even if $A$ is uncountable? @JohnGriffin – sequence Sep 22 '17 at 01:25
  • @sequence No. If $A$ is infinite, then $|\mathcal{F}(A)|=|A|$. – John Griffin Sep 22 '17 at 01:27
  • So if $A$ is uncountable and $B_a$ is finite, then how does this imply that $\bigcup B_a$ is countable? @JohnGriffin – sequence Sep 22 '17 at 01:28
  • @sequence It doesn't. We only know that $|\cup_{a\in A}B_a|\le|A|$. – John Griffin Sep 22 '17 at 01:33
  • You say that we know that $B=|\bigcup B_a|\le |\mathbb{N}|$. But how? Since if $B$ is uncountable (since $A$ is uncountable and there are uncountably many $a\in A$), its cardinality may be greater than $\aleph_0$. @JohnGriffin – sequence Sep 22 '17 at 02:40
  • @sequence I didn't say that anywhere. We know $B=\cup_{a\in A}B_a$ because $B$ is linearly independent and $A$ is spanning. We know $|\cup_{a\in A}B_a|\le|A||\mathbb{N}|$ by the lemma and the fact that each $B_a$ is finite. Finally $A$ infinite implies $|A||\mathbb{N}|=|A|$. – John Griffin Sep 22 '17 at 02:44
  • By the lemma, we can say that $|\bigcup B_a|\le |A||B|$, so why $|A||\mathbb{N}$|? @JohnGriffin – sequence Sep 22 '17 at 03:04
  • @sequence That's an unfortunate use of notation on my part. The $B$ in the lemma is not the same as the $B$ in the proof which follows. In the lemma, $B$ is a set whose cardinality is at least that of every $B_a$. In the proof, this role is filled by $\mathbb{N}$. I'll edit the answer to fix this issue. – John Griffin Sep 22 '17 at 03:07