Consider the natural bijection between $(0,1]$ and the infinite $\{0,1\}$-sequences with infinitely many ones (written in base $2$). 


Define also the sequence $(x\_n)$ by $x\_n=1$ if $n$ is odd and $x\_n=0$ otherwise. Hence, for each $\omega \in (0,1]$, we can consider the subsequence $((x \upharpoonright \omega)\_n):=(x\_{n\_k})$, where the index $n\_k$ is taken iff there is a $1$ in the corresponding representation of $\omega$.


Lastly, let $\nu\_2(n)$ be the $2$-adic valuation of a positive integer $n$, i.e., the biggest integer $k\ge 0$ such that $2^k$ divides $n$.



> 
> **Question.** Is it true that 
>  $$
> S:=\left\{\omega \in (0,1]: \left\{k: (x \upharpoonright \omega)\_n=1, \nu\_2(n)=k \text{ for infinitely many }n\right\}\text{is finite}\}\right\}
> $$ 
>  is a first Baire category set?
> 
> 
> 



Meager set of subsequences
The answer is yes.


Denote here by $\mathbb{N}$ the set of positive (nonzero) integers. First consider the application :
$$f : (y\_k)\_{k \in \mathbb{N}} \longrightarrow \sum\_{k \in \mathbb{N}} 2^{-(y\_1 + \ldots + y\_k)}.$$
This is a continuous bijection from the Baire space $\mathbb{N}^\mathbb{N}$ to $(0, 1]$ (to a sequence $(y\_k)$, we associate the real number whose dyadic expansion has 1's exactly in the positions $y\_1$, $y\_1 + y\_2$, $y\_1 + y\_2 + y\_3$,...) Moreover, the direct image by $f$ of an open set has nonempty interior, so it is easy to verify that the direct image of a nowhere dense set is nowhere dense, and so that the image of a meager set is meager. Hence, it is enough to verify that $T := f^{-1}(S)$ is meager.


For $y = (y\_n)\_{n \in \mathbb{N}} \in \mathbb{N}^\mathbb{N}$, we have $(x\restriction f(y))\_m = 1 \Leftrightarrow y\_1 + \ldots + y\_m \text{ is odd}$. So we have that $y \in T$ iff $\{k \in \mathbb{N} \mid y\_1 + \ldots + y\_{2^k(2n + 1)} \text{ is odd for infinitely many n}\}$ is finite, what is equivalent to say that $\exists k\_0 \, \forall k \geqslant k\_0\, \exists n\_0 \, \forall n\geqslant n\_0 \: (y\_1 + \ldots + y\_{2^k(2n + 1)} \text{ is even})$. In particular, $T$ is Borel so has the Baire property. To prove that it is meager, we cannot apply the topological 0-1 law but we will imitate its proof. Suppose that $T$ non-meager, and consider $s$ such that $T$ is comeager in $N\_s$, the basic open set associated to $s$.


For each permutation $\sigma : \mathbb{N} \longrightarrow \mathbb{N}$ with finite support (i.e. $supp(\sigma) = \{n \in \mathbb{N} \mid \sigma(n) \neq n\}$ is finite), we associate an automorphism $u\_\sigma$ of the Baire space, defined by $u\_\sigma(y) = (y\_{\sigma(n)})\_{n\in\mathbb{N}}$. The set $T$ is invariant under all these automorphisms, since for $n, k >\max(supp(\sigma))$, $u\_\sigma(y)\_1 + \ldots + u\_\sigma(y)\_{2^k(2n + 1)}$ and $y\_1 + \ldots + y\_{2^k(2n + 1)}$ have the same parity. Hence, $T$ is comeager in $U := \bigcup\_\sigma u\_\sigma(N\_s)$, where the union is indexed by all the $\sigma$ with finite support. But $U$ is a dense open set: indeed, if we consider a basic open set $N\_t$, then we have that $N\_{s \frown t} \subseteq N\_s$, so taking a $\sigma$ such that $u\_\sigma(N\_{s \frown t}) = N\_{t \frown s}$, we get that $N\_{t \frown s} \subseteq U$, so $N\_t \cap U \neq \varnothing$. We finally deduce that $T$ is comeager.


But now consider the automorphism $v$ of the Baire space defined by $v(y)\_n = y\_n$ for $n \geqslant 2$, $v(y)\_1 = 2m + 1$ if $y\_1 = 2m$, and $v(y)\_1 = 2m$ if $y\_1 = 2m + 1$. Then for every $m \geqslant 1$, $v(y)\_1 + \ldots + v(y)\_{2^k(2n + 1)}$ and $y\_1 + \ldots + y\_{2^k(2n + 1)}$ don't have the same parity, so $v(T) \cap T = \varnothing$. But both $T$ and $v(T)$ are comeager, so this is a contradiction.


