functional-analysislinear algebra
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.
You have to be careful with your wording. Saying "a span is finite" doesn't really mean anything. The span of a collection of vectors is the set of all finite linear combinations of those vectors.
Consider the vector space of all real polynomials $\mathcal{P}(\mathbb{R})$. It has a basis $\{x^n \mid n \in \mathbb{N}\cup\{0\}\}$ which has infinite cardinality, so $\mathcal{P}(\mathbb{R})$ is infinite dimensional. Any finite linear combination of these polynomials gives you an element of $\mathcal{P}(\mathbb{R})$. If you were to take a finite collection of the polynomials, say $\{x^{n_i} \mid i =1, \dots, k\}$, then their span would not contain any polynomial of degree more than $\max_i n_i$, so you would not obtain $\mathcal{P}(\mathbb{R})$.
It doesn't work. See Keith Conrad's note (namely page 16). Here is a relevant screenshot.
Best Answer
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)