# A vector space is either finite-dimensional or infinite-dimensional and cannot be both

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 $$i \iff \overline{f}$$

• You need at least the axiom of countable choice, see this question. – user10354138 Jun 9 at 16:24
• @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? – Tom Jun 9 at 17:32
• 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. – user10354138 Jun 9 at 17:53
• Thanks a lot and thanks for the reference: I did not know this result had a name. – 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.
• 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? – Tom Jun 9 at 17:38
• Ops you re right! What about the other direction, i.e. that $i$ implies $\overline{f}$? – Tom Jun 9 at 17:44