The product of locally compact spaces is locally compact

My definition of local compactness is as follows:

A topological space $$(X, \mathcal{T})$$ is locally compact if, for every $$x \in X$$, there is a $$U \in \mathcal{T}$$ with $$x \in U$$ such that $$\overline{U}$$ is compact.

I am then asked to prove that, if $$X$$ and $$Y$$ are locally compact and Hausdorff, then $$X \times Y$$ equipped with the product topology is also locally compact. Trouble is, I don't believe I use Hausdorff in my proof! And I don't see where it breaks down. Here it is:

Let $$(X, \mathcal{T}_X)$$ and $$(Y, \mathcal{T}_Y)$$ be locally compact. Consider a point $$(x, y) \in X \times Y$$. Then there exist $$U \in \mathcal{T}_X$$ with $$x \in U$$ and $$V \in \mathcal{T}_Y$$ with $$y \in V$$ such that $$\overline{U}$$ is compact in $$X$$ and $$\overline{V}$$ is compact in $$Y$$.

Firstly, $$U \times V$$ is open in $$X \times Y$$, and $$(x,y) \in U \times V$$. Furthermore, $$\overline{U}$$ is compact as a subset of $$X$$, so considered as a topological space with the subspace topology, it is also compact. Similarly for $$\overline{V}$$. Then the space $$\overline{U} \times \overline{V}$$ equipped with the product topology is also compact. Now I believe that the product topology induced by the topologies on $$\overline{U}$$ and $$\overline{V}$$ is the same as the subspace topology induced by considering $$\overline{U} \times \overline{V}$$ as a subspace of $$X \times Y$$. So $$\overline{U} \times \overline{V}$$ is also compact with respect to the subspace topology, and so it is a compact subset of $$X \times Y$$. But $$\overline{U} \times \overline{V} = \overline{U \times V}$$, so we're done.

• @mathworker21 (I wrote "I believe" since I haven't seen a proof of that fact, but I have tried proving it myself, and I think my proof is correct. So I didn't want to claim it as definitely true!) Well, the problem specifically asks me to assume Hausdorff, but other than that I don't have any reason to believe so. It also didn't make sense to me intuitive why I would need it, so I tried proving it without that assumption. – Danny Hansen Nov 24 '18 at 14:36
• ok, I guess if it said "locally compact and Hausdorff" that is weird. usually things will say "locally compact Hausdorff" or just assume everything is Hausdorff just for ease. sorry for my first comment. – mathworker21 Nov 24 '18 at 14:39
• @mathworker21 Well, it's not in English, but directly translated it does say "locally compact Hausdorff" (it says "assume $X$ and $Y$ are locally compact Hausdorff spaces"). But unless I am mistaken that just means that the space is both locally compact and Hausdorff. Sorry if that makes a difference. Local compactness is only defined in the context of this problem, so I don't know any results about it. – Danny Hansen Nov 24 '18 at 14:42
• It doesn't make a real difference. What I was trying to say is that people usually assume things are Hausdorff, just to sometimes make their life easier. When they do this, they say "locally compact Hausdorff". But if someone said "locally compact and Hausdorff", I would think they were emphasizing the word "Hausdorff", which would be weird. If what I'm saying sounds nonsensical to you, it's because it mostly is. Goodnight – mathworker21 Nov 24 '18 at 14:46
• A brief comment: to say that a space $X$ is locally foo generally means that given any point $x \in X$ in the space and any neighborhood $U$ of $x$, there is a neighborhood $V$ of $x$ with $V \subseteq U$ and having the property foo. In the presence of the Hausdorff condition, your definition is equivalent to what I would therefore call locally compact. – Stephen Nov 24 '18 at 19:04

$$\overline{U\times V} = \overline{U} \times \overline{V}$$ in the product topology and the product of compact sets is compact, in short.