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I need to show that: If $X$ is locally compact, second countable and Hausdorff, then $X^*$ is metrizable and hence $X$ is metrizable.

I have already showed that every locally compact Hausdorff space is regular. Then I use that $X$ is regular and second countable to apply Urysohn's metrization theorem and say that $X$ must be metrizable. However, this doesn't show that $X^*$ (the one point compactification) is metrizable. I know that $X^*$ is compact and regular, but I don't know that $X^*$ is second countable. How can I show that $X^*$ is second countable and hence metrizable? Or, if it is not possible, how else should I approach this problem?

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Since $X$ is second countable, all you need for $X^*$ to be second countable is a countable neighborhood base for the point at infinity, right? Since $X$ is locally compact and second countable, you can cover $X$ with countably many open sets whose closures are compact. Then every compact subset of $X$ is contained in a finite union of those open sets, and the complements of the closures of those unions will provide a countable neighborhood base for $\infty$.

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Try to establish the following facts:

  • $X$ is $\sigma$-compact.
  • $X$ is hemicompact.
  • Use the hemicompactness of $X$ to construct a countable neighborhood base at $\infty$ in $X^*$.
  • Conclude that $X^*$ is second countable.

See Theorem 3.44 in Aliprantis–Border (2006, pp. 92–93) for more details.

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