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?


1 Answer 1


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.


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