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It would be very appreciated if someone could review my solution. Thanks!

Problem:

Show that every locally compact Hausdorff space is completely regular.

Proof:

Let $X$ be a locally compact Hausdorff space. One point sets are closed in $X$ since $X$ is Hausdorff.

Then there is a space $Y$ such that $Y$ is the one point compactification of $X$ where $Y$ is compact Hausdorff.

Let $x_0 \in X$ and $B \subset X$ such that $x_0 \notin B$ and $B$ is closed in $X$. Then the open set $U = X - B$ contains $x_0$. Then since $U$ is open in $X$ it is also open in $Y$ by construction of the one point compactification Munkres gives. Then $C = Y - U$ is a closed set in $Y$ that contains $B$.

Then since Y is compact Hausdorff it is also normal. Hence since the set $\{x_0\}$ is closed since Y is Hausdorff, by the Urysohn lemma we can define a continuous function $f: Y \to [a,b]$ such that $f(x_0) = a$ and $f(x) = b$ for every $x \in C$.

But since $f$ is continuous we can take the continuous function $g$ on a subspace of the domain of $f$. Namely, $g: X \to [a,b]$ since $X$ is a subspace of $Y$. Hence we have a continuous function $g$ that maps $g(x_0) = a$ and $g(B) = b$, since $B \subset C$. We can replace $[a,b]$ by $[0,1]$ to satisfy the definition of completely regular.

Hence $X$ is completely regular.

Martin Sleziak
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H_1317
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2 Answers2

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It follows from two basic facts:

  1. $X$ embeds into its one-point compactification $\alpha X$, which is Hausdorff and compact and thus normal and hence Tychonoff.

  2. A subspace of a Tychonoff space is again Tychonoff.

Your proof is almost a re-proof of the last fact. If it has been covered in your course, it's better to just quote that instead of redoing its proof in a special case.

Henno Brandsma
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  • Ok so looks like tychonoff space is another word for a completely regular space from googling around -- munkres doesn't use that term. He does prove a subspace of a completely regular space is completely regular, so I could've used that..... With that said, is my proof correct regardless? – H_1317 Apr 17 '19 at 02:04
  • @H_1317 yes, it works. But it’s better to use the things you already know. – Henno Brandsma Apr 17 '19 at 03:40
  • Do you happen to know of any bibliography that clearly states "locally compact Hausdorff spaces are completely regular"? – André Caldas Aug 16 '22 at 14:28
  • https://topology.pi-base.org/theorems/T000027/references cites Theorem 19.3 in Willard's General Topology (which I confirmed). – Steven Clontz Nov 17 '23 at 17:29
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You can prove it easy using this properties:

  • $X$ Locally compact Hausdorff, then exists $X^{\ast}$ one point compactification
  • $X$ is subspace of $X^{\ast}$
  • $X^{\ast}$ is Hausdorff compact, then $X^{\ast}$ is Normal.
  • Normal implies completely regular
  • Subspace of completely regular is completely regular
Christian Coronel
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