Is the Baire space $\sigma$-compact?

The Baire space is the set $\mathbb{N}^\mathbb{N}$ of all sequences of natural numbers under the product topology taking $\mathbb{N}$ to be discrete. It is a complete metric space, for example with the metric $d ( x , y ) = \frac{1}{n+1}$ where $n$ is least such that $x(n) \neq y(n)$.

A topological space $X$ is called $\sigma$-compact if it is the countable union of compact subsets.


$\omega^\omega$ is not σ-compact.

First note that every compact subset of $\omega^\omega$ has empty interior; that is, they are nowhere dense. (If $K \subseteq \omega^\omega$ is compact with nonempty interior, then there is a finite sequence $s = (\ell_0, \ldots , \ell_n )$ in $\omega$ such that $[s] = \{ x \in \omega^\omega : x\text{ extends }s \}$ is a subset of $K$. As $[s]$ is clopen it must be compact itself. However we can write $[s]$ as $\prod_{i \in \omega} A_i$ where $A_i = \{ \ell_i \}$ for $i \leq n$, and $A_i = \omega$ for $i > n$, and since $\omega$ is not compact it follows that $[s]$ cannot be compact, a contradiction!)

As a complete metric space (hence a Baire space), the Baire Category Theorem then implies that the Baire space cannot be σ-compact.

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    $\begingroup$ Here is another way of proving this: The continuous image of any $\sigma$-compact set is $\sigma$-compact. However, any analytic set is the continuous image of Baire space. $\endgroup$ – Andrés E. Caicedo Mar 20 '14 at 5:40

$\omega^\omega$ is not $\sigma$-compact. This is a diagonal-style argument: Suppose that $\omega^\omega = \bigcup_n K_n$ where the $K_n$ are compact. For each fixed $n$, for every projection $\pi_m: \omega^\omega \to \omega$ the set $\pi_m[K_n]$ is compact hence there is some $N(n,m) \in \omega$ such that $\pi_m[K_n] \subseteq [0,N(n,m)]$.

Now the point $p \in \omega^\omega$ defined by $p_m = N(m,m)+1$ does not lie in any $K_n$, for if $p \in K_m$ for some $m$, then $p_m \in \pi_m[K_m] \subseteq [0, N(m,m)]$, while $p_m$ is one larger than that.


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