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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

Martin Sleziak
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Learnmore
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  • Haven't read carefully, but http://math.stackexchange.com/questions/219441/compactification-of-non-compact-metrizable-space?rq=1 – Silvia Ghinassi Feb 10 '16 at 17:21
  • Yes and Yes. For 1., show that $Y$ has a countable basis. – Daniel Fischer Feb 10 '16 at 17:23
  • Also related: http://math.stackexchange.com/questions/62820/metrizable-compactifications – Asaf Karagila Feb 10 '16 at 17:26
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    @AsafKaragila;I don't know how the link you shared is related to the above as there are many concepts both in the question and the answer such as Polish space which I am not accustomed to – Learnmore Feb 11 '16 at 02:21
  • for 1) you could also look at the family of continuous functions having limit at infinity and take the Hausdorff compactification relative to such family, which is metrisable since the family is separable – Tommaso Seneci Apr 28 '21 at 16:37

1 Answers1

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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.

Henno Brandsma
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  • 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? – Learnmore Feb 11 '16 at 02:26
  • How did you know that $ \overline {\mathcal 0}$ is compact ? – Learnmore Feb 11 '16 at 02:27
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    @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). – Henno Brandsma Feb 13 '16 at 13:41
  • @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). – Henno Brandsma Feb 13 '16 at 13:43
  • That was my question how do you have a base of open sets such that $\overline O$ is compact? – Learnmore Feb 14 '16 at 18:17
  • @learnmore It follows directly from local compactness. What's your definition then? – Henno Brandsma Feb 14 '16 at 18:18
  • I mean that there exists a compact set $C$ containing an open set $O$ around $x$ – Learnmore Feb 14 '16 at 18:25
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    Fix $x$ and pick $C_x$ compact, $O_x$ open such that $x \in O_x \subseteq C_x$. Then all open $O$ subsets of $O_x$ have the property that $\overline{O}$ has compact closure (as it's closed in the compact $C_x$). So every point has a local base of such "closure is compact" sets. – Henno Brandsma Feb 14 '16 at 19:09