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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?

Idonknow
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1 Answers1

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$\Bbb N^{\Bbb N}$ 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.
Brian M. Scott
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  • +1. Very nice and instructive answer.I guessed correctly that $N^N$ had the product topology-it's hard to see what other topology would be relevant in this case. – Mathemagician1234 Mar 18 '15 at 06:33