How do I show these properties of maps between vector spaces relate to the properties of linear independence/spanning? Consider $V\neq 0$, a Vector Space over a field $F$. Let $S \subset V$ be a non-empty set. Consider the following properties of $S$:
(1) For any Vector Space $W$ over $F$, any map $f : S \to W$ extends to a linear map from $V$ to $W$.
(2) For any Vector Space $W$ over $F$ any two linear maps $f,g : V \to W$ satisfying $f(s) = g(s)$ for all $s \in S$, we have $f(v)=g(v)$ $\forall v \in V$.
(3) $S$ is linearly independent.
(4) The span of $S$ is $V$.
Then prove that (1) implies (3) and (2) implies (4).
MY ATTEMPT:
I am aware of the fact that every linearly independent set containing finite list of vectors can be extended to form a basis of a finite dimensional vector space $V$. But this result might fail to give me a conclusion here, since the vector space can contain infinite list of vectors as well! Also there exist another thought as if I consider any arbitrary set $S$ then the choices of it being linearly dependent or independent are equivalent, at the same time it need not be necessary that Span of $S$ will be $V$. Any help?
Thanks in advance :)
 A: Suppose $S$ is linearly dependent. That is, there exists some $x \in S$, $x_1, \ldots, x_n \in S \setminus \{x\}$, and $\alpha_1, \ldots, \alpha_n \in F$ such that
$$x = \alpha_1 x_1 + \ldots + \alpha_n x_n.$$
Consider a map $f : S \to F$ that maps everything to $0$, except $x$, which it maps to $1$. Suppose $T : V \to F$ is a linear map extending $f$. Then,
$$1 = Tx = \alpha_1 Tx_1 + \ldots + \alpha_n Tx_n = 0,$$
a contradiction. Thus, no such $T$ exists, so $\lnot (3) \implies \lnot(1)$, i.e. $(1) \implies (3)$.
Proving $(2) \implies (4)$ is where it would be handy to assume finite-dimensions, or at least assume you were more comfortable with infinite-dimensions. It's not harder, as all the standard results still hold, they just require different proofs, and the knowledge of how bases work in infinite dimensions. Essentially, we argue like this:

*

*If $S$ does not span, then $\exists v \in V \setminus \operatorname{span} S$.

*Reduce $S$ to a (Hamel) basis $B$ of $\operatorname{span} S$ (This still works; you need Zorn's lemma).

*Show $B \cup \{v\}$ is linearly independent.

*Extend $B \cup \{v\}$ to a (Hamel) basis $C$ of $V$ (more Zorn's lemma).

*Take the map $f : C \to F$ that maps everything to $0$, except $v$ which it maps to $1$, and extend it linearly to $V$ (you can do this uniquely with Hamel bases, just like in finite-dimensions, but once again, you need Zorn's lemma).

*Such a map agrees with the $0$ map on $\operatorname{span} B = \operatorname{span} S$, and hence agrees on $S$, but is not the same as the $0$ map, due to what it does to $C$. Thus, $\lnot (4) \implies \lnot (2)$.

I'm reluctant to go into more detail than this, because I don't want to re-explain the theory to you from the perspective of infinite-dimensions. If you're happy with finite-dimensions, then my outline above still works, and should work within the framework of finite-dimensions that you've (hopefully) covered so far.
