Meager set from randomly signing a convergent series

Question

Given a convergent series of real numbers $\sum_{n}a_n$, consider the set $$X=\left\{\sigma:\Bbb N\to \{-1,1\}\mid \mbox{$\sum$}_n \sigma(n)\cdot a_n\text{ converges}\right\}$$ as a subset of the Cantor space $\{-1,1\}^{\Bbb N}$. Then $X$ has full Lebesgue measure iff $\sum_n {a_n}^2$ converges.

I'm looking for a similar result for Baire category. Particularly, which condition is sufficient for $X$ to be meager?

Answer 1

Turns out that $X$ is comeager when $\sum_n |a_n|<\infty$ (which is obvious), and meager otherwise.

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Suppose $\sum_n |a_n|=\infty$. For each $k\in\Bbb N$ let $f_k:\{-1,1\}{}^\Bbb N\to \Bbb R$ send $\sigma\mapsto \sum_{0\leq n\leq k} \sigma(n)\cdot a_n$.

Given $p>0$, let $A(p)=\{\sigma\in\{-1,1\}^\Bbb N\mid \forall k(| f_k(\sigma)|\leq p)\}$, then $A(p)$ is closed:

let $\tau\notin A(p)$, then $|f_k(\tau)|>p$ for some $k$, and let $U_{\tau\restriction k}=\{\sigma\in{}^\omega\{-1,1\}\mid \tau\restriction k\subset \sigma\}$, then $U_{\tau\restriction k}$ is an open set that separates $\tau$ from $A(p)$.

Moreover, $A(p)$ is nowhere dense:

Take an arbitrary basic nonempty open $U_s=\{\sigma\in\{-1,1\}^\Bbb N\mid s\subset \sigma\}$ with $s\in\{-1,1\}^m$ for some $m\in\Bbb N$. Since $\sum_n |a_n|=\infty$, by making enough terms positive, there exists $t\in \{-1,1\}^{m'}$ for $m'>m$ such that $s\subset t$ and $\sum_{0\leq n\leq m'} t(n)\cdot a_n>p$, so then $U_t\cap A(p)=\varnothing$ and $U_t\subset U_s$.

Hence $\bigcup_{p\in \Bbb N}A(p)$ is meager.

If $\sigma\in X$, then $\sum_n\sigma(n)\cdot a_n$ converges, and thus is bounded. Therefore $\sigma\in A(p)$ for some large enough $p\in\Bbb N$, showing that $X\subset \bigcup_{p\in\Bbb N}A(p)$. Therefore $X$ is meager.

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Based on a similar proof for subseries, so where $\sigma:\Bbb N\to\{0,1\}$, by Tibor Šalát, found in "On subseries of divergent series", Mat. Casopis. Sloven. Akad. Vied 18 (1968)