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Suppose I have a topological vector space $X$ and a second category set $B$ in that space. Is it the case that $B$ contains a basis for $X$? For instance, in normed linear spaces, an open ball (which is second category) surely contains a Hamel basis for the space because all possible 'directions' are contained in it. I was wondering whether this fact can be extended to both topological vector spaces from NLS as well as to second category sets from open balls. In case we cannot obtain a Hamel basis, can we get a Schauder basis?

Not Euler
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    What sort of basis do you have in mind? See What is the difference between a Hamel basis and a Schauder basis? Hamel basis does not depend on topology and for a Schauder basis the space has to be at least first countable. – Conifold Feb 29 '20 at 01:30
  • @Conifold Yes I do know the difference. I was hoping to get the whole picture; that's why I didn't mention which basis. Do you have an example in mind where the Hamel basis or the Schauder basis isn't contained in such a set? Would love to hear about it. – Not Euler Feb 29 '20 at 01:38
  • Please make the body of your question self-contained rather than just giving essential information in the title only. When you fix this it will be much easier to discuss your question. – Rob Arthan Feb 29 '20 at 01:42

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No. For instance, let $X$ be any separable infinite-dimensional Banach space. In particular, $X$ is second-countable so it has only $2^{\aleph_0}$ different closed sets and thus only $2^{\aleph_0}$ different countable unions of closed sets with empty interior. Let $(S_\alpha)_{\alpha\in\mathfrak{c}}$ be an enumeration of all such countable unions, so every first category subset of $X$ is contained in some $S_\alpha$. Fix some nonzero vector $x\in X$; we can then by transfinite recursion construct a sequence $(b_\alpha)_{\alpha<\mathfrak{c}}$ such that $\{x\}\cup\{b_\alpha\}_{\alpha<\mathfrak{c}}$ is linearly independent and $b_\alpha\not\in S_\alpha$ for each $\alpha$. (At each step, the set of choices of $b_\alpha$ which would maintain the linear independence is the complement of a proper linear subspace of $X$, which cannot have first category since $X$ is covered by two of its translates, and thus is not contained in $S_\alpha$.) The set $B=\{b_\alpha\}_{\alpha<\mathfrak{c}}$ then has second category in $X$ by construction, but does not contain a Hamel basis for $X$ since its span does not contain $x$. (And if $X$ itself has no Schauder basis, then $B$ cannot contain one either.)

Eric Wofsey
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    Wonderful proof. I was wondering whether I can also make this set $B$ to be not dense in $X$. For that I picked a ball $G$ of radius $r$. Consider any proper linear subspace $V$ of $X$. This time around I want to pick elements from $A=V^c\cap G^c$ so that my $B$ does not intersect $G$. I need only prove $A$ is second category in $X$. If I translate by $2r$ units in any direction, the translated ball and the original ball do not overlap. Thus when I take the union of $A$ and its $2r$-translate, what I'm (possibly) missing is $1$st category in $X$. – Not Euler Mar 01 '20 at 06:32
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    Sure, that would work. Though at the end, it's not that what you're missing is 1st category; rather, what you're missing is contained in $V$, so you know its complement is 2nd category. – Eric Wofsey Mar 01 '20 at 16:04
  • Also, what's the exact reason for considering a second countable space? What would go wrong if the number of first category sets exceeded the continuum? – Not Euler Mar 02 '20 at 18:32
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    You need to know that the number of $S_\alpha$'s is at most the (algebraic) dimension of $X$, so that there is no risk that ${x}\cup{b_\beta}_{\beta<\alpha}$ will already span all of $X$ at some stage. – Eric Wofsey Mar 02 '20 at 18:43