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I am trying to prove the following statement

Let $V$ be vector space over an infinite field $F$, and $S$ be an at most countable subset of $V$. Then $|L(S)| = |F|$, where $|S|$ denotes the cardinality of $S$ and $L(S)$ is the linear span of $S$.

Can you please help me to construct a bijection between $L(S)$ and $F$, or give me another idea?

Anne Bauval
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  • $S$ must contain at least one nonzero vector. Then, first prove that $|L(S)|\ge|F|.$ – Anne Bauval Sep 26 '23 at 05:13
  • For the reverse inequality, use that ($\forall n\in\Bbb N$) $|F^n|=|F|$ (this requires the axiom of choice if I remember well) – Anne Bauval Sep 26 '23 at 05:19
  • @AnneBauval could you perhaps give some indication of why you think this needs axiom of choice (or perhaps a name of something that I can look up) because it seems to me at least that one does not need axiom of choice for sets that are countable or have a cardinality of the continuum, so it would be interesting to see if/why it breaks down – Carlyle Sep 26 '23 at 05:45
  • $|A^2|=|A|$ is easy to prove indeed without AC when $A=\Bbb N$ or $\Bbb R.$ For the general case I do not "think" AC is necessary. I only vaguely remember. @Carlyle – Anne Bauval Sep 26 '23 at 05:52
  • @Carlyle I found this: https://math.stackexchange.com/questions/1617795/a2-a-for-every-infinite-a-iff-axiom-of-choice-holds?noredirect=1 – Anne Bauval Sep 26 '23 at 06:05
  • You can prove the statement without constructing an explicit bijection. You do need to assume that $S \neq { 0 }$, but with that assumption, it's quite easy to show that $\vert L(S) \vert \geq \vert F \vert$. Then use the fact that $S$ is at most countable to prove the other inequality. – Robert Shore Sep 26 '23 at 06:27
  • @AnneBauval thanks! – Carlyle Sep 26 '23 at 06:50
  • Robert's comment was more or less included in my 2 first ones above. In the first one: not only do you need to assume that $S \neq { 0 }.$ It must also be $\ne\varnothing.$ – Anne Bauval Sep 26 '23 at 07:07

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Note that: $$L(S)=\{\sum_{i=1}^n a_iv_i:v_i\in S,a_i\in F,n\in\mathbb{N}\}=\bigcup_{n\in\mathbb{N}} \{\sum_{i=1}^n a_iv_i:v_i\in S,a_i\in F\}=\bigcup_{n\in\mathbb{N}}A_n$$

For each $A_n$, you have a surjection $S^n\times F^n\to A_n$ given by: $$(v_1,\ldots,v_n,a_1,\ldots,a_n)\to\sum_{i=1}^n a_iv_i$$ So $|A_n|\leq |S^n\times F^n|$. It is clear that also $|F|\leq |A_n|$. As $F$ is infinite and $S$ at most countable: $$|F|\leq |A_n|\leq |S^n\times F^n|=\max\{|S^n|,|F^n|\}=|F^n|=|F|$$ Therefore $|A_n|=|F|$. A countable union of such sets again has the same cardinality as $F$, since $F$ is infinite.

Math101
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  • Your map $A_n\to S^n\times F^n$ is not well defined, and there is no obvious bijection between these 2 sets. The best you can do is define a surjection in the other direction. – Anne Bauval Sep 26 '23 at 05:32
  • You can moreover freely improve and include my 2 comments above. – Anne Bauval Sep 26 '23 at 05:34
  • @AnneBauval Fixed it. For some reason I thought the question said $S$ is a linearly independent set. – Math101 Sep 26 '23 at 05:46
  • Even so, there wouldn't have been any obvious bijection between these 2 sets. + There remain things to exploit in my 2 initial comments above. – Anne Bauval Sep 26 '23 at 05:58