Any $G \subseteq \mathbb{N}^\mathbb{N}$ is clopen

Question

I have a question concerning the space $\mathbb{N}^\mathbb{N}$.

I found in Srivastava's book on Borel sets that the sets of the form $$\Sigma (s) := \{ \alpha \in \mathbb{N}^\mathbb{N} \ | \ s \prec \alpha, s \in \mathbb{N}^{<\mathbb{N}} \}$$ are clopen, where $\mathbb{N}$ is endowed with the discrete topology, and $\mathbb{N}^\mathbb{N}$ is endowed with the product topology.

Now, I think that any $G \subseteq \mathbb{N}^\mathbb{N}$ is clopen.

"Proof:" By taking an arbitrary $n \in \mathbb{N}$ as an index for the projection function, $\pi_n (G) \subseteq \mathbb{N}$ is open, because $\mathbb{N}$ is endowed with the discrete topology. The same line of reasoning is applied to show that they are closed.

Questions:

1. Is this correct?
2. Is it possible to prove that sets $\Sigma (s)$ are open through metric arguments (i.e. neighborhoods,$\varepsilon$)?

As always, any feedback is most welcome.
Thank you for your time.

Answer 1

That the sets $\Sigma(s)$ are open is immediate from the definition of the product topology on $\Bbb N^{\Bbb N}$, so we need only show that they are closed. Let $s\in\Bbb N^{<\Bbb N}$; then $s\in\Bbb N^n$ for some $n\in\Bbb N$.

• Show that $\Sigma(s)=\Bbb N^{\Bbb N}\setminus\bigcup\big\{\Sigma(t):t\in\Bbb N^n\setminus\{s\}\big\}$; then $\Sigma(s)$, being the complement of an open set, must be closed.

It is definitely not true that every subset of $\Bbb N^{\Bbb N}$ is clopen; I’ll construct a fairly simple set $A\subseteq\Bbb N^{\Bbb N}$ that is neither open nor closed.

For each $n\in\Bbb N$ let $x^{(n)}=\left\langle x_k^{(n)}:k\in\Bbb N\right\rangle\in\Bbb N^{\Bbb N}$ be defined by

$$x_k^{(n)}=\begin{cases} 1,&\text{if }k\le n\\ 0,&\text{if }k>n\;. \end{cases}$$

Let $A=\left\{x^{(n)}:n\in\Bbb N\right\}$.

• Show that $A$ is not open in $\Bbb N^{\Bbb N}$. Probably the easiest way to do this is to use the fact that the sets $\Sigma(s)$ defined in your question form a base for the topology on $\Bbb N^{\Bbb N}$ and show that for each $s\in\Bbb N^{<\Bbb N}$, $\Sigma(s)\nsubseteq A$.

• Show that $A$ is not closed in $\Bbb N^{\Bbb N}$ by showing that the constant sequence $\langle 1,1,1,\ldots\rangle$, which is not in $A$, is nevertheless in the closure of $A$.