Let $V$ be a vector space over the field $\mathbb{F}$ and let's define the properties $(f)$ and $(i)$ in the following way:

$f$) $\exists F \in V$ finite such that $\text{span}(A) = V$

$i$) $\exists I \in V$ infinite such that $I$ is lineary independent

If I now define a vector space to be finite-dimensional if it satisifes $(f)$ and infinite dimensional if it satifies $(i)$, I definitely need to make sure that every vector space is either finite-dimensional or infinite-dimensional and cannot be both.

So the question is how to prove that $$ f \iff \overline{i} $$ or equivalently that $$ i \iff \overline{f} $$

  • $\begingroup$ You need at least the axiom of countable choice, see this question. $\endgroup$ – user10354138 Jun 9 at 16:24
  • $\begingroup$ @user10354138 Thx. So from what I understands from Asaf Karagila's answer to that question, you can prove (assuming the axiom of countable choice) that $\overline{f} \implies i$. And what about the converse? $\endgroup$ – Tom Jun 9 at 17:32
  • $\begingroup$ If a finite set $F$ spans the space, you cannot find more than $\#F$ linearly independent vectors. This is part of the Steinitz exchange lemma. $\endgroup$ – user10354138 Jun 9 at 17:53
  • $\begingroup$ Thanks a lot and thanks for the reference: I did not know this result had a name. $\endgroup$ – Tom Jun 9 at 18:00

Hint: if (f) fails, you can inductively construct a linearly independent sequence $x_1, x_2, \dots$.

If $x_1, \dots, x_{n-1}$ have been chosen, then by assumption their span is not equal to $V$, so you can choose an $x_n$ which is not in their span...

Conversely, suppose $f$ holds, so that there is a finite set, call it $F = \{x_1, \dots, x_n\}$ which spans $V$. If $I$ is an infinite set, you can find $n+1$ distinct elements $y_1, \dots, y_{n+1}$ in it. By elementary linear algebra, they cannot be linearly independent.

  • $\begingroup$ thx actually this (as does the answer to this math.stackexchange.com/questions/300494/…) shows that $\overline{f}$ implies something even stronger than $i$, namely the existence of a countably infinite linearly independent set. Right? $\endgroup$ – Tom Jun 9 at 17:38
  • $\begingroup$ Yes. But assuming some fairly weak version of the axiom of choice (which is already used in this argument), then every infinite set has a countably infinite subset, so you automatically get your "stronger" conclusion anyway. (And the proof of the latter is basically the same as what I wrote: given an infinite set, start picking elements one by one, and you will never run out...) $\endgroup$ – Nate Eldredge Jun 9 at 17:41
  • $\begingroup$ Ops you re right! What about the other direction, i.e. that $i$ implies $\overline{f}$? $\endgroup$ – Tom Jun 9 at 17:44
  • $\begingroup$ @Tom: See edit. $\endgroup$ – Nate Eldredge Jun 9 at 17:52
  • $\begingroup$ Thank you Nate - now it s all clear $\endgroup$ – Tom Jun 9 at 17:59

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