Under what conditions does a metrizable space have a metrizable compactification?
Solution
A metrizable space \(X\) has a metrizable compactification if and only if it is second countable. For metrizable spaces this is equivalent to separability.
Suppose first that \(X\) is dense in a compact metrizable space \(Y\). Every compact metrizable space is second countable, and every subspace of a second-countable space is second countable. Hence \(X\) is second countable.
Conversely, suppose \(X\) is metrizable and second countable. A metrizable space is regular, so by the Urysohn metrization/embedding theorem in its standard embedding form, a second-countable regular space admits a topological embedding \[ e:X\hookrightarrow [0,1]^{\mathbb Z_+}. \] The Hilbert cube \([0,1]^{\mathbb Z_+}\) is compact metrizable. Therefore \[ Y=\overline{e(X)} \] is a closed subspace of a compact metrizable space, hence is itself compact metrizable, and \(e(X)\) is dense in \(Y\). Thus \(Y\) is a metrizable compactification of \(X\).
Hence \[ \boxed{X\text{ metrizable has a metrizable compactification}\iff X\text{ is second countable}.} \]