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Let $V$ be a (Hausdorff) topological vector space, and let $S = \{v_i\} _{i \in \mathbb N} \subset V$ be a generating subset (notice that it is countable!).

Does countability make it possible to prove the existence of a Schauder basis $B \subseteq S$ (contained in $S$) without Zorn's lemma?

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    $\begingroup$ Generally, $V$ need not admit a Schauder basis. If $V$ admits a Schauder basis, I would still expect that we can have generating sets (that means $\operatorname{span} S$ is dense in $V$, I presume?) which don't contain a Schauder basis. I don't see an obvious example, however. $\endgroup$ – Daniel Fischer Dec 3 '17 at 21:49
  • $\begingroup$ Another example: $\mathscr{O}(\mathbb{D})$ (the space of holomorphic functions on the unit disk) with the generating set $\{ 1 + z^k : k \in \mathbb{N}\setminus \{0\}\}$ which doesn't contain a Schauder basis. $\endgroup$ – Daniel Fischer Dec 4 '17 at 11:36
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With or without Zorn's lemma, one cannot construct a Schauder basis without detailed knowledge of the space. It is not a thing that drops out from some abstract argument. There are topological vector spaces, even Banach space, with no Schauder basis at all.

Even if the space $V$ has a Schauder basis, we shouldn't expect to obtain it from some set of vectors with dense linear span. Being a Schauder basis requires much more than being a linearly independent set with dense linear span. One has to be able to represent every vector in the space as the limit of partial sums of a series $\sum c_n x_n$, not just by some arbitrary sequence of linear combinations of $c_n$.

A standard example: The set of monomials $S = \{x^n : n=0,1,2,\dots\}$ has dense linear span in $C[-1,1]$ (the space of continuous functions on $[-1,1]$ with uniform norm), because every $f\in C[-1,1]$ can be uniformly approximated by polynomials (Weierstrass' theorem). It is also linearly independent, since the only polynomial that is identically $0$ is the polynomial with zero coefficients. But $S$ is not a Schauder basis, because the continuous function $f(x) = |x|$ does not admit a representation $f(x) = \sum_{n=0}^\infty c_n x^n$. Indeed, the functions with a power series representation are infinitely differentiable on $(-1,1)$, where one can differentiate the power series term by term.

The above example also answers your question in the negative. No subset of $S$ is a Schauder basis, since throwing out some monomials will not help us represent the function $f(x)=|x|$ as a series.

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  • $\begingroup$ Very clear answer. Do linearly independent subsets with dense span have a name? It seems that I can obtain these using Zorn. (Ah, and good to see good old NormalHuman back!) $\endgroup$ – Alex M. Dec 4 '17 at 13:11
  • $\begingroup$ Not that I know of. "A linearly independent subset with dense span" is what I would call it. $\endgroup$ – user357151 Dec 4 '17 at 13:12

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