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$(X,\tau)$ is a local compact, second countable Hausdorff space s.t. $ X = \cup_n K_n$ for countably many compact sets in $X$. Then $\infty$ has a countable neighborhood basis of open sets where $\infty \in (X_\infty,\tau_\infty)$. This is the one-point compactification of $X$.

This is homework so I would appreciate some hints.

I already saw that the assumption that $X = \cup_n K_n$ is redundant since $X$ is locally compact en second countable. However if $\mathcal N$ is an open neighborhood basis for $\infty$, I want to show: If $U = X_\infty \setminus K$ is a open neighborhood which contains $\infty$ we want to find a set $X_\infty \setminus K' \in \mathcal N$ s.t. $X_\infty \setminus K' \subset X_\infty \setminus K$ ,i.e. $K \subset K'$.

José Carlos Santos
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  • It is immediate from the def'n of open nbhd base that if $N$ is an open nbhd base at $\infty$ and if $U$ is an open set containing $\infty$ then there exists $V\in N$ with $\infty \in V\subset U.$ That is, $K'= X$ \ $V\supset X$ \ $U=K.$ – DanielWainfleet Jul 23 '17 at 03:34

1 Answers1

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Do it in steps:

  • First show that there is a cover $U_1\subseteq\overline{U_1}⊆U_2⊆\overline{U_2}⊆...$ where all $U_i$ are open and all $\overline{U_i}$ are compact. Start with $U_0=\emptyset$. Now assume by induction that $U_n$ with compact $\overline{U_n}$ has been defined. For each point in $\overline{U_n}\cup K_{n+1}$ there is an open neighborhood with a compact closure. Since $\overline{U_n}\cup K_{n+1}$ is compact, finitely many of them cover this set. What can you say about the closure of their union?
  • Once you have proven the existence of such a sequence, consider an open neighorhood $V$ around $\infty$. Its complement is compact and covered by the $U_n$. Can you find a larger compact set?
Stefan Hamcke
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  • Thank you. This is very helpful, except that every point has an open neighborhood with compact closure. But I will figure this out, I guess ;D –  Nov 24 '13 at 18:27