Let $X$ be a locally compact Hausdorff space.Let $Y$ be the one-point compactification of $X$. Two questions are:

  1. Is it true that if $X$ has a countable basis then $Y$ is metrizable?
  2. Is it true that if $Y$ is metrizable then $X$ has a countable basis?

My attempt:We know that every compact space which is metrizable has a countable basis.Thus in (2) we have $Y$ is $2^{nd}$ countable and a subspace of a $2^{nd}$ countable space being $2^{nd}$ countable so $X$ is $2^{nd}$ countable .

In (1) I could only figure out that $X$ is regular since it is locally compact Hausdorff space.Also $X$ has a countable basis so by Urysohn Metrization Theorem $X$ is metrizable.

But how can this help me conclude whether $Y$ is metrizable/not?

Any help will be helpful


The argument for 2 is correct.

For 1, you can show that $Y$ has a countable base as well: as $X$ is locally compact and second countable, it has a countable open base $\mathcal{B}$ such that $\overline{O}$ is compact for all $O \in \mathcal{B}$.

Then the point at infinity $\infty$ has a local base of the form $\{\infty\} \cup \{X \setminus C: C = \cup_{i=1}^n \overline{O_i}, O_i \in \mathcal{B}\}$, which is countable (little argument requireed). Show it is a local base for $\infty$ (which uses that $X$ is lcoally compact), and combine it with $\mathcal{B}$ to form a countable base for the whole compact Hausdorff $Y$, which is then metrisable by Urysohn again.

  • $\begingroup$ I can't figure out that if I show even that $Y$ is second countable(which I could not show) how does that show that $Y$ is metrizable;Can you kindly elabarote? $\endgroup$ – Learnmore Feb 11 '16 at 2:26
  • $\begingroup$ How did you know that $ \overline {\mathcal 0}$ is compact ? $\endgroup$ – Learnmore Feb 11 '16 at 2:27
  • $\begingroup$ @learnmore You do quote Urysohn's metrisation theorem: a regular $T_1$ second countable space is metrisable. So if you show $Y$ is second countable you are done (as it is already compact Hausdorff). $\endgroup$ – Henno Brandsma Feb 13 '16 at 13:41
  • $\begingroup$ @learnmore In a locally compact Hausdorff space you always have a base of open sets such that $\overline{O}$ is compact. In a second countable space, you can reduce that base to a countable one. That's the base I start with. So I pick them with compact closures (as we are in a locally compact space I can do that). $\endgroup$ – Henno Brandsma Feb 13 '16 at 13:43
  • $\begingroup$ That was my question how do you have a base of open sets such that $\overline O$ is compact? $\endgroup$ – Learnmore Feb 14 '16 at 18:17

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