Every linearly independent set can be extended to a basis Let $E$ be  linear space (infinite dimensional in general). We know by Zorn's lemma that there exists a basis. Now let $S \subset E$ be any linear independent subset. How to prove that it is contained in some basis of $E$? 
And moreover if $F \subset E$ is subspace then there are linear functional such that $f(F) = 0$ and linear complement to $F$ in $E$.
I know how it can be done for finite dimensional spaces but I am always confused when infinite dimension and Zorn's lemma are involved.
 A: Use again Zorn's Lemma. Define:
$$C:=\left\{\,T\supset S\,/\,\,T\;\text{is linearly independent}\,\right\}$$
Observe that $\;S\in C\implies C\neq\emptyset\;$ and we can partial order $\;C\;$ by set inclusion and etc. (check this!), so by Zorn's Lemma we're done if we succeed in proving a maximal element $\;M\;$ in  $\;C\;$ is a basis of $\;E\;$ , but this is easy as otherwise there'd be an element $\;x\in E\,,\,\,x\notin\text{ Span}\,M\;$, but then $\;M\cup\{x\}\;$ is lin. ind. and contains $\;S\;$ so contradiction...(fill in details here)
A: Let $U$ be the span of $S$. Then the vector space $E/U$ has a basis, say $T=\{y_i+U:i\in I\}$.
Consider $S'=S\cup\{y_i:i\in I\}$ and prove it is a basis for $E$.
The span of $\{y_i:i\in I\}$ is a complement for $U$. A linear functional that is $0$ on $U$ can be found by just choosing an element $y_i$ and decreeing that this one is mapped to $1$ and all other elements of $S'$ are mapped to $0$ (provided $S$ is not a basis of $E$ to begin with, in this case the zero functional is the only possible one).
