# Clopen subsets of $A^\Bbb N$ for finite $A$

Let $A$ be a finite set with the discrete topology and let $X = A^\Bbb N$ be the product space. Let $$\pi_n:X\to A^n$$ be the projection map, i.e. $\pi_n(x_1,\dots,x_{n},x_{n+1},\dots) = (x_1,\dots,x_n)$. Each such map induces a partition of $X$ which we denote as $[x]_n=:\pi_n^{-1}(\pi_n(x))$. Such sets $[x]_n$ are the cylinders of the product topology on $X$. As it has been pointed out here, for each $x$ and $n$ the cylinder $[x]_n$ is clopen in $X$ and hence those are finite unions of cylinders.

My question is the following: is it true that clopen subsets of $X$ are exactly finite unions of cylinders - otherwise I am interested in a counterexample. Does the situation change in case $A$ is countable?

I guess this question is also related, but I am not sure whether answers there apply directly here.

-

Yes.

Let $\sigma \in A^{<\mathbb{N}}$, i.e. a finite length string of elements of $A$. Let $[\sigma] = \{f \in A^{\mathbb{N}} : \sigma \preceq f\}$. $[\sigma]$ is a clopen set. So finite union of these cyclinder sets are clopen.

By Tychonoff Theorem, this space is compact. (You can also use Weak König Lemma to prove compactness.) Suppose $L$ is a clopen subset of $X = A^{\mathbb{N}}$. The cylinders form a basis for this topology. $L$ is open so it is a countable union $\mathcal{A}$ of cyclinders. Since $L$ is closed subset of a compact set, $L$ is compact. So a finite subcollection of $\mathcal{A}$ of these cyclinders cover $L$. But the union of the cylinders in $\mathcal{A}$ is $L$, so these finitely many cylinders also union up to $L$. So $L$ is a finite union of cyclinders.

-
Thank you for the answer - I have some questions, if I may. As far as I understand, here we use the fact that each clopen set is compact and is open. Thus, we can cover it with elements of the basis and there is a finite subcover. Very nice and neat! Just a few comments: 1. with $\sigma \preceq f$ do you mean that $f$ is a obtained from $\sigma$ by concatenation? 2. we can cover $L$ with a countable number of cylinders not because its open, but because there are only countably many them, right? 3. Due to the nature of the proof, I guess it does not work when $A$ is countable, does it? –  S.D. Dec 3 '12 at 14:07
If the length of $\sigma$ is $n$, then $\sigma \preceq f$ means $\sigma(i) = f(i)$ for all $i < n$, i.e. $\sigma$ is an initial segment of $f$. You can cover with countable many cylinders because the cylinders form a countable basis for $X$. Even when $A$ is countable, $A^\mathbb{N}$ has a countable basis. The part that fails is that $A^\mathbb{N}$ is not longer compact. –  William Dec 3 '12 at 16:56
Thanks for the clarification –  S.D. Dec 3 '12 at 16:59
so would it be right to say, that any clopen subset of a compact topological space is a finite union of its basic elements? –  S.D. Dec 4 '12 at 7:08

Just to add a bit about $A^\mathbb{N}$ where $A$ is infinite, consider just $\mathbb{N}^\mathbb{N}$.

Note that $$U = \bigcup_{i \in \mathbb{N}} [ \langle 2i \rangle ]$$ is a union of infinitely many cylinders whose complement $$V = \bigcup_{i \in \mathbb{N}} [ \langle 2i+1 \rangle ]$$ is also a union of infinitely many cylinders. Therefore both of these are clopen. So the situation does change in the infinite case.

Of course, you can copy examples of infinite unions of cylinders in $\{0,1\}^\mathbb{N}$ which are not clopen in this case, as well. For example, if for each $i \in \mathbb{N}$ we denote by $\sigma_i$ the sequence $\langle 0 \cdots 0 1 \rangle$ of length $i+1$, then while $$W = \bigcup_{i \in \mathbb{N}} [ \sigma_i ]$$ is clearly open, it is not closed as the constant zero sequence is in the closure of the set.

-
Could you tell me, what is meant by $[\langle 2i\rangle]$? –  S.D. Dec 3 '12 at 16:50
@S.D. $\langle 2i \rangle$ denotes the length $1$ sequence whose only entry in $2i$, and $[ \sigma ]$ denotes, as you have done above, the cylinder above $\sigma$ for $\sigma$ a finite sequence. –  Arthur Fischer Dec 3 '12 at 16:59