TITLE: Why all vector space have a span set?
QUESTION [6 upvotes]: I thought about this question, but I don't sure if my proof is correct. In the book, he put this question like a observation of span sets' definition, so I tried proof this.
My attempt:
Suppose that exists a vector space $V$ such that there isn't a span set $S = \{ v_1, v_2, \ldots, v_n \}$, so exists a $v_{n+1} \in V$ such that $v_{n+1} \neq \sum_{i=1}^{i=n} v_i$, so $S \cup \{ v_{n+1} \}$ can be a span set of $V$, but can be exists $v_{n+2} \in V$ such that $v_{n+2} \neq \sum_{i=1}^{i=n+1},  v_i$, so $S \cup \{ v_{n+1}, v_{n+2} \}$ can be a span set of $V$ and we can be in this cycle infinitely. My doubt is what ensures that doesn't exists infinitely many vectors that can be span by a span set?

REPLY [8 votes]: Your argument doesn't work - there are indeed vector spaces which have no finite spanning set (an interesting example of this is $\mathbb{R}$ as a $\mathbb{Q}$-vector space). However, there is a much simpler proof: if $V$ is a vector space, then $V$ is a spanning set of itself.
If you want a basis - that is, a spanning set which is linearly independent - then things are much trickier. In fact, without the axiom of choice, they need not exist! We can however construct a basis for $V$ if the axiom of choice holds, using either Zorn's Lemma or transfinite induction (they're ultimately the same, just packaged differently).