Existence of increasing sequence of compact sets

Question

Let X be a complete separable metric space. Can we always find an increasing sequence of compact sets $K_j$ such that $X=\cup K_j$?

In $\mathbb{R}$ this is immediate, for example taking $K_j=[-j,j]$. This also extends to higher dimensions. In fact, we also have the added benefit that the Lebesgue measure of the boundary of $K_j$ is $0$. Can we conclude the same for X (wrt some finite Borel measure)?

Answer 1

This is not possible in any infinite dimensional separable Banach space, in particular $C[0,1]$. Proof: use Baire Category Theorem to conclude that one of the $K_n$'s has non-empty interior. This implies that the space is finite dimensional.

Answer 2

Another example: $X=\mathbb{N}^\mathbb{N}$ in the product metric (when $\Bbb N$ has the discrete metric) is complete metric. (it's homeomorphic to the irrationals $\Bbb P$ BTW), but no countable union of compact subsets can cover $X$.

This is a classical diagonal argument: let $K_m, m \in \Bbb N$ be a sequence of compact subsets of $X$. Projections are continuous so all $\pi_n[K_m], n \in \Bbb N$ are finite subsets (the only compact subsets of $\Bbb N$ are finite sets), and now define a point $x = (x_n)_n \in X$ by

$$\forall n: x_n = \max(\pi_n[K_n])+1 \text { or } x_n = \min(\Bbb N \setminus \pi_n[K_n])$$

Then $x \notin \cup_m K_m$, for suppose it does: $x \in K_p$ for some $p \in \Bbb N$. But $x \in K_p$ implies $x_p \in \pi_p[X_p]$ but $x_p \notin \pi_p[K_p]$ by construction. Contradiction.