A number $x \in (0,1]$ written in non-terminating dyadic expansion $$ \sum_{n\ge 1} \frac{e_n(x)}{2^n} $$ is said to be normal if $\frac{1}{n}(e_1(x)+\cdots+e_n(x))\to \frac{1}{2}$ as $n\to \infty$. Let $N$ be the set of normal numbers.

It is known that $N$ is a set of measure one and, at the same time, a meager set (see the answers of Andreas Blass here and here). However, there is something which I don't understand in the proof of the meagerness of $N$:

Proof. I take for granted that the space $\{0,1\}^{\mathbf{N}}$ is completely metrizable (with the product topology of the discrete topology on $\{0,1\}$). Hence we want to prove, equivalently, that the set of sequences $$ (e_1(x),e_2(x),\ldots) \in \{0,1\}^{\mathbf{N}} $$ such that $x \in N$ is meager. Baire's category applies, hence it is enough to show $N^c$ contains a countable intersection of open dense sets (which is a $G_\delta$ dense set).

At this point, we have to construct this $G_\delta$ set, let us $$ G_\delta=\bigcap_{n\ge 1}G_n $$ with each $G_n$ open and dense in $\{0,1\}^{\mathbf{N}}$. I recall that the topology on $\{0,1\}^{\mathbf{N}}$ is zero-dimensional, i.e., has a base of clopen sets. The suggestion in the first linked thread is to take something like $$ G_n=\{(e_1(x),e_2(x),\ldots) \in \{0,1\}^{\mathbf{N}}: e_k(x)=\cdots=e_{2^k}(x)=0\} $$ where $k=k(n)$ is a sufficiently large integer (let us say greater than $k(n-1)$). It is also easily seen that $$ N\cap G_\delta=\emptyset\,\,\Leftrightarrow\,\,N\subseteq \left(\bigcap G_n\right)^c \,\,\Leftrightarrow\,\, N^c \supseteq \bigcap G_n. $$ The conclusion would follow from the fact that $G_n$ is open and dense. But $G_n$ is a member of the base of the product topology of $\{0,1\}^{\mathbf{N}}$, i.e., it is clopen (in particular, its closure coincides with the same set, hence it cannot be dense!).

I suppose that I should change a bit the definition of each $G_n$ so that the final inclusion still holds and, at the same time, each $G_n$ is open and dense. How I should modify it?

(Note that Andreas Blass writes: "$G_n$ doesn't contain all such points $x$; some endpoints have to be removed to make $G_n$ open" but I cannot see how this suggestion could work: if $G_n^\prime$ is a subset of $G_n$ then the closure of $G_n^\prime$ is contained in $G_n$, hence $G_n^\prime$ cannot be dense too.)

Edit. Probably I found a way: define $$ \bar{G}_\delta:=\bigcap_n \bar{G}_n,\,\,\,\text{where} \,\,\,\bar{G}_n:=\bigcup_{i\ge n}G_i $$ for each $n$. Then each $\bar{G}_n$ is open and $$ \bar{G}_\delta=\bigcap_n \bigcup_{i\ge n}G_i=\{e(x) \in \{0,1\}^{\mathbf{N}}: e(x) \in G_i \,\text{ infinitely often}\}. $$ In particular, the above inclusion still holds. We miss only to prove that each $\bar{G}_n$ is dense. Let $F$ be the closure of $\bar{G}_n$, we want to prove that $F=\{0,1\}^{\mathbf{N}}$. Equivalently, it is claimed that $$ \mathrm{Int}\left(\bar{G}_n^c\right)=\emptyset. $$ If this is not true, then there exists a nonempty clopen set (in the base of the product topology) which is contained in $$ \bar{G}_n^c=\bigcap_{i\ge n}G_n^c. $$ After a moment thought (and considering that here that are infinitely many constraints), this is impossible.

In particular, this proves that the set of $x \in (0,1]$ such that $$ \liminf_{n\to \infty}\frac{e_1(x)+\cdots+e_n(x)}{n}>0 $$ is meager.

However, this is far from the small parenthesis of Andreas Blass: does there exist a quicker way to obtain the same conclusion?

  • $\begingroup$ The terminology I am familiar with is that a normal number is normal in all integer bases, not just base 2. $\endgroup$ – Asaf Karagila Jun 29 '17 at 13:28
  • $\begingroup$ It depends on the context: I saw many times the definition you are referring to. This one can be found, e.g., in Billingsley, Probability and Measure, p. 8 $\endgroup$ – Paolo Leonetti Jun 29 '17 at 14:42
  • $\begingroup$ The set that you call $\tilde G_n$ here is the set that I called $G_n$ in the first of the two answers that you linked to. That it is dense is easy to see: Given any nonempty open interval $I$, start with any point $x\in I$; find some $k>n$ such that, if you modify the binary digits of $x$ after the $k$-th one, then the result is still in $I$ (such a $k$ exists as $I$ is open); and then do such a modification, substituting 0 for a lot of digits after the $k$-th one, to obtain a point in $G_n$. $\endgroup$ – Andreas Blass Jun 29 '17 at 14:51

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