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There is a theorem which says that if $V$ is a vector space over a field $\mathbb{K}$ and $V$ admits an infinite linearly independent set, then any two bases $B$ and $B'$ have the same cardinality. So I wanted to prove this theorem. The question is followed by a hint which says that the following lemma can be applied to prove this theorem.

Lemma: Let $A\ne \emptyset \ne B$ be two sets. A map $\phi: B\to C$ is countable-to-one if for every $c\in C$, $\phi^{-1}(\{c\})$ is countable. If such a map exists then $|B| \le \aleph_0|C|$.

The idea was to apply this lemma to the collections of finite sets of $B$ and $B'$, i.e. $\mathcal{F}(B)$ and $\mathcal{F}(B')$, which is quite straightforward, except that $\mathcal{F}(B)$ may well be uncountable.

So I asked my professor for a hint and here's what he said:

Given $E\in\mathcal{F}(B)$, there is a minimal finite set $f(E)$ of elements of $B'$ for which $E$ is contained in span$(f(E))$.

He also added that this is a simple fact from linear algebra. But I didn't recognize this fact. Can someone please explain the concept from linear algebra above in simpler terms?

Namely, am I correct in re-interpreting this statement as follows?:

Given $E\in\mathcal{F}(B)$, there is a bijective function $f$ on $E$ which takes elements of $B$ into elements of $B'$ such that $E$ is spanned by $f(E)$ (that is, by a minimal finite subset of $B'$).

Anyway, this statement doesn't seem to be a trivial concept from linear algebra, and I don't think I've ever heard about it.

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The trivial fact that your professor alluded too is basically the definition of the span of an infinite set of vectors. If $B$ is an infinite set of vectors then $\text{span} \ B$ is by definition the set of vectors that can be written as a linear combination of some finite subset of vectors in $B$.

The span of a basis is the entire space so given any one vector $v \in B$ there is a finite subset of $B'$ such that $v$ is in the span of that subset. Thus given any finite subset $E \in \mathcal F(B)$ there is a finite subset of $B'$ whose span contains $E$.

As for your re-interpretation: If you think about the one element subsets of $B$ your re-interpretation seems to say that every element of $B$ is a scalar multiple of an element in $B'$, and this is not necessarily true so your re-interpretation looks incorrect to me.

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  • $\begingroup$ Do you mean it is necessarily true that if $B$ is an infinite set of vectors then it can be spanned by a finite basis? Then, I think, infinite bases would not exist. $\endgroup$ – sequence Sep 23 '17 at 0:44
  • $\begingroup$ No, this is only true for finite sets of vectors. $\endgroup$ – Jim Sep 23 '17 at 1:41
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    $\begingroup$ I did not say that. I said that for each vector $v$ in the span of $B$ there is a finite set of vectors from $B$ such that $v$ is a linear combination of the vectors in this finite set, but I did not say that you can use the same finite set of vectors each time. $\endgroup$ – Jim Sep 24 '17 at 3:27
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    $\begingroup$ $\mathbb R$ does not span $\mathbb Q$, it's not even contained in $\mathbb Q$. Generally if $V$ is a vector space over the coefficient field $k$ and $S$ is a set of vectors in $V$ then $\text{span}_k(S)$ is the subspace of $V$ whose elements are vectors that can be written in the form $\sum_{i = 1}^na_iv_i$ for some choice of a positive integer $n$, $n$ coefficients $a_i \in k$, and $n$ vectors $v_i \in S$. It $\endgroup$ – Jim Sep 27 '17 at 22:56
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    $\begingroup$ And as for infinitely many rationals that add to a real, yes you can write a real number as a limit of an infinite sequence of sums of rational numbers, but the definition of a span in linear algebra is that you are only allowed to have finitely many vectors. $\endgroup$ – Jim Sep 27 '17 at 23:07

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