# Is the Baire space $\sigma$-compact?

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!)
• 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. – 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.