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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.

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    What do you mean by "...and linear complement to F in E" ? – 2017-01-16
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    Subspace $L$ such that $E$ can be written as direct sum of $F$ and $L$. – 2017-01-16
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    @Vl So you want to say that any subspace of $\;E\;$ has a complement...but then what is the role of that linear functional $\;f\;$ in this?? – 2017-01-16

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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)

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    Can I now choose the basis $x_{\nu}$ of $F$, extend it to the basis of $E$ by what you just proved and then define $f: E \to C$ ( $C$ is a field) as $f(y) = 1$ where $y$ is an element of basis which is not one of $x_{\nu}$ ? – 2017-01-16
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    @Vladislav Yes, I think you can do that now....and zero elsewhere, of course. – 2017-01-16
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    it would be zero elsewere anyway 'cause kernel has codimension 1. – 2017-01-16
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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).