Metrizable compactifications

Solution 1:

First of all, a compact metric space is second countable, hence a metrizable space can be homeomorphic to a subspace of a compact metrizable space only if it is second countable.

So let's assume $X$ is a separable metrizable space. Choose a compatible metric $0 \leq d \leq 1$, and complete $X$ to get a Polish space $\overline{X}$ with a homeomorphic copy of $X$ inside. Now, as I argued in my answer here, every Polish space is homeomorphic to a $G_{\delta}$ inside the Hilbert cube $[0,1]^{\mathbb{N}}$.

The embedding itself is easy, simply choose a dense subset $(x_n)_{n \in \mathbb{N}} \subset X$ and map $x$ to $(d(x,x_n))_{n \in \mathbb{N}} \in [0,1]^{\mathbb{N}}$ (recall that we chose a bounded metric $0 \leq d \leq 1$). This is obviously continuous, and it is not hard to show that it's a homeomorphism onto its image. To see that the image of $\overline{X}$ is a $G_{\delta}$ is harder and given in detail in the answer to Apostolos's question I mentioned above.

Upshot: every separable metrizable space is homeomorphic to a subspace of the Hilbert cube (this one of the 100 variants and refinements of the Urysohn theorem).

Note:

• The one-point compactification is not a viable option, as it is Hausdorff only if $\overline{X}$ is locally compact.
• We can't do better than a $G_{\delta}$ for complete spaces, since open subsets of (locally) compact spaces are locally compact, hence in order to have a compactification in the stricter sense that $\overline{X}$ be open and dense, local compactness of $\overline{X}$ is necessary.

Finally, if a locally compact space is metrizable, then it is necessarily second countable, hence its one-point compactification is second countable as well, and, again by Urysohn metrizable and $\overline{X}$ is an open subset of its one-point compactification.

For more on this, consult Kechris, Classical descriptive set theory, Springer GTM 156, Springer 1994. See in particular Theorem 5.3 on page 29.