0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.