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

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.

• No, this is not correct. The fact that every $\pi_n(G)$ is open does not mean that $G$ is open (even in the Euclidean plane, there are subsets all of whose projections are open and which are not open!, like the disjoint union of an open annulus with a point inside). – Gro-Tsen May 11 '16 at 15:09
• Thanks for the feedback. I will go back to work on it. – Kolmin May 11 '16 at 15:29
• Another idea: the sets $\Sigma(s)$ form a base of the topology; but not all open sets are closed, since the latter are not “closed” under arbitrary unions (no pun intended). – Pedro Sánchez Terraf May 11 '16 at 15:51

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 $A$.

• Thanks a lot: as always, great (really great!) answer! Just two numbered questions below. – Kolmin May 11 '16 at 21:39
• 1) Question about the (not completely clear to me) fact that the openness of $\Sigma (s)$ in $\mathbb{N}^{\mathbb{N}}$ comes from the definition of product. Does this come from the fact that essentially $\Sigma (s)$ is constrained only on the elements in common with $s$ (that are open, because they are singletons in the discrete topology), and then takes any value in $\mathbb{N}$, for every other $n \in \mathbb{N}$ outside $s$? Now I see why this should also be the reason why these sets form a basis for the topology. – Kolmin May 11 '16 at 21:45
• 2) I was proceeding with a proof along metric arguments, for two reasons: (a) I blindly didn't see what you pointed out; (b) I thought it was good exercise. Is such a proof actually doable? My line of reasoning was to use the tree metric between elements in $\Sigma (s)$, and its complements (by using the distance function). – Kolmin May 11 '16 at 21:50
• @Kolmin: You’re welcome! Yes, if $s\in\Bbb N^n$, then $\Sigma(s)$ restricts precisely the first $n$ coordinates, each to an open singleton, and makes no restriction on any other coordinate, so it’s a basic open set in the usual base for the product topology. And it turns out that these sets are already enough to give us a base for the product topology. \\ Yes, you could carry out a proof using a metric. How easy it would be would depend on which metric you chose; taking $d(x,y)=2^{-n}$ for $x\ne y$, where $n$ is minimal such that $x_n\ne y_n$, is probably the easiest. But you end up ... – Brian M. Scott May 11 '16 at 22:20
• ... essentially repeating the non-metric argument, since it turns out that the open balls are basically the sets $\Sigma(s)$. – Brian M. Scott May 11 '16 at 22:21