Compact set is nowhere dense in $\mathbb{N}^{\mathbb{N}}$

Question

Show that any compact set is nowhere dense in $\mathbb{N}^{\mathbb{N}}$, the set of all infinite sequences.

A set $A$ is nowhere dense if the interior of its closure is empty, i.e. int$(\bar{A})=\emptyset$.

I have no idea how to start to show the statement. Can anyone give some hints?

Answer 1

$\Bbb N^{\Bbb N}$ is Hausdorff, so every compact subset of $\Bbb N^{\Bbb N}$ is closed. Thus, you need only show that no compact subset of $\Bbb N^{\Bbb N}$ contains a non-empty open set. For this it’s helpful to know a nice base for the topology on $\Bbb N^{\Bbb N}$.

For each finite sequence $\sigma=\langle a_0,a_2,\ldots,a_{n-1}\rangle$ of natural numbers let

$$B(\sigma)=\left\{x\in\Bbb N^{\Bbb N}:x_k=a_k\text{ for }k<n\right\}\;,$$

and let

$$\mathscr{B}=\bigcup_{n\in\Bbb N}\left\{B(\sigma):\sigma\in\Bbb N^n\right\}$$

be the family of all such sets.

• Show that $\mathscr{B}$ is a base for the product topology on $\Bbb N^{\Bbb N}$.
• Show that each $B(\sigma)\in\mathscr{B}$ is closed as well as open.
• Show that each $B(\sigma)\in\mathscr{B}$ is homeomorphic to $\Bbb N^{\Bbb N}$, which is not compact.
• Conclude that no compact subset of $\Bbb N^{\Bbb N}$ can contain any member of $\mathscr{B}$, and hence that every compact subset of $\Bbb N^{\Bbb N}$ must have empty interior.