# Alexandroff compactification question

If $X$ is a locally compact and metrizable space such that its Alexandroff compactification is not first countable. Does this imply that no other compactification of $X$ can be first countable? Why?

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The only extra info on $X$ you have is that it is not separable. Any metrizable $X$ has a metrizable compactification iff it is separable (for $X$ locally compact the Aleksandrov one will do, non-separable spaces will have others). So the question comes down to: does a non-separable metrizable space have a first countable compactification? –  Henno Brandsma Dec 3 '10 at 20:19

This follows from a theorem of Banakh and Leiderman in "Uniform Eberlein compactifications of metrizable spaces" which states that a metrizable space $X$ has a first-countable uniform Eberlein compactification if and only if $|X|\leq\mathfrak{c}$.
This implies that the discrete space on $\omega_1$ (which is trivially locally-compact and metrizable) has a first-countable compactification. But the Alexandroff compactification is not first-countable, since the open subsets of the "point at infinity" consist of all co-finite subsets of $\omega_1$ and any (sub-)basis must be uncountable.