# Locally compact Hausdorff then Tychonoff

This is my proof. If $$X$$ is locally compact Hausdorff then for each $$x \in X$$ its compact neighborhoods form a neighborhood basis at $$x$$. For an arbitrary closed set $$C \not\ni x$$ let $$V \subset X \setminus C$$ be a compact neighborhood of $$x$$. For each dyadic rational $$r$$, let's define open neighborhoods $$U_r$$ inductively s.t. $$s < r \implies \overline{U_s} \subset U_r$$. Let $$U_0 = \operatorname{int}(V), U_1 = X - C$$. $$\{A \mid A \subset U_1 \text{is a compact neighborhood of a }x \in V\}$$ is a cover of $$V$$ so there is a finite subcover $$S$$. Let $$W = \bigcup{S}$$ then it is compact because it is a finite union of compact sets. Let $$U_{1/2} = \operatorname{int}(W)$$. Then it is open and $$\overline{U_0} \subset U_{1/2}, \overline{U_{1/2}} \subset U_1$$. Apply this method analogously to define for all other dyadic rationals. Now let's define a function $$f\colon X \to [0, 1]$$ $$f(x) = \begin{cases} \operatorname{inf}\{r \mid x \in U_r\}& x \in U_1 \\ 1& x \not\in U_1 \end{cases}$$ Then $$f(x) = 0$$ and $$f \equiv 1$$ on $$C$$. Let's show $$f^{-1}(\infty, a)$$ and $$f^{-1}(b, \infty)$$ are open. If $$a > 1$$ it is $$X$$ and if $$a \leq 1$$, $$f^{-1}(\infty, a) = \bigcup{\{U_r \mid r < a\}}$$ so open. $$f^{-1}(b, \infty) = \bigcup{\{X \setminus U_r \mid r > b\}} = \bigcup{\{X \setminus \overline{U_s} \mid s > b\}}$$ so open. Finally $$f$$ is continuous so $$X$$ is Tychonoff. Is this proof correct?

## 1 Answer

It is possibly correct (you do essentially use normality like structures we get from the local compactness), but why not use the fact that the Alexandroff (one-point) compactification $$\alpha(X)$$ of $$X$$ is compact Hausdorff (and so is normal and hence Tychonoff) and contains $$X$$ as a subspace (and subspaces of Tychonoff spaces are Tychonoff)? Much cleaner. Why redo the proof of Urysohn's lemma for $$X$$ when we already have it for $$\alpha(X)$$? Efficiency.