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$\def\R{{\mathbb R}}$ May I please receive help with the following problem? It is from Munkrees Topology Textbook.

We know the projection to $X$ or $Y$ of an open subset of $X\times Y$ with the product topology is necessarily an open set.

(ii) Prove $X$ and $Y$ are compact topological spaces, then the projection of a closed set of $X\times Y$ to $Y$ is a closed set.

$\textbf{Solution:}$ If $\mu \subseteq X \times Y$ is closed, it must also be compact because closed sets in compact spaces are compact. Being the image of a compact set, $p(\mu)$ is compact. Because $p(\mu) \subseteq Y$ is Hausdorff, and compact sets of Hausdorff spaces are closed, $p(\mu)$ must be closed.

rudinsimons12
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  • These look good. Except: (ii) did not say Hausdorff. (Unless you have Bourbaki definitions, where compact includes Hausdorff.) But you say $Y$ is Hausdorff. Why? – GEdgar Mar 26 '20 at 15:14
  • @GEdgar I was not sure about $(ii)$, I took a guess because I was thinking we can still find disjoint open sets in Y? i am not sure. – rudinsimons12 Mar 26 '20 at 15:26
  • "Assume any topological space is both Hausdorff and non-empty, and any examples given must also be Hausdorff and non-empty." –  Mar 26 '20 at 15:27
  • You need not assume Hausdorffness at all for (ii). – Henno Brandsma Mar 26 '20 at 16:27

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As others in the comments have said, your example for i. is good. I'd just be careful when you say

$A$ is closed, and $p_2(A) = \mathbb{R} \setminus \{0\}$ is open (not closed)

This could be construed to mean that because $p_2(A)$ is open, it is not closed. It's possible for sets to be both open and closed, so you may want to re-write it (this is very nitpicky).

Your proof of ii. is good, if a bit unclear. Because you say you're not sure about it, I'll spell it out in more detail with links to relevant theorems.

If $\mu \subseteq X \times Y$ is closed, it must also be compact, because closed sets in compact spaces are compact. Being the image of a compact set, $p(\mu)$ is compact. Because $p(\mu) \subseteq Y$ is hausdorff, and compact sets of hausdorff spaces are closed, $p(\mu)$ must be closed.

Noah Caplinger
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  • Thank you Noah, I made the change to my solution and just removed the part where i say (not closed), would that be fine? – rudinsimons12 Mar 26 '20 at 16:20
  • Here, the relevant detail is that $p_2(A)$ is not closed (that's what the question is asking), so you would replace "is open (not closed)" with "is not closed." As I said, this is a tiny thing. Anyone reading this proof understands what you're going for. – Noah Caplinger Mar 26 '20 at 16:36
  • I see, thank you, Noah. For part ii, I was given a link by. Henno, and I was wondering if what I added to my solution for ii.. is valid? Because I am still a bit confused. – rudinsimons12 Mar 30 '20 at 20:36
  • There are two proofs going on here: 1. yours (and mine) which assume that $Y$ is hausdorff. With this assumption, your proof (just the first paragraph after "Solution:") is correct. 2. You actually don't need the hausdorff condition (as Henno notes) this gives a slightly different proof. – Noah Caplinger Mar 30 '20 at 20:45
  • Thanks, so what you provided and what I added for the ii is fine now? I deleted the two additional paragraphs – rudinsimons12 Mar 30 '20 at 20:53
  • if you are assuming the spaces are hausdorff, that proof is fine. However, I notice that you've dropped that assumption ("Assume any topological space is both Hausdorff and non-empty") from the question, which now makes your proof incorrect (with the given assumptions). Hopefully the situation is clear. – Noah Caplinger Mar 30 '20 at 22:57
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The idea of (i) is fine, your proof of why $A$ is closed needs work. For points in the plane with $x \neq 0$ your proof works to show we have an open neighbourhood (in $\Bbb R^\ast \times \Bbb R$ which is itself open in the plane so the neighbourhood is open in the plane too) that misses $A$, but for points with $x=0$ we need a separate argument. A closed set of $\Bbb R^\ast \times \Bbb R$ need not be closed in the plane, in general!

For (ii) This is true without Hausdorffness of any of the spaces while your proof assumes Hausdorffness in the codomain. The projection along a compact space is always closed and this characterises compactness; proofs can be found here, e.g.

Henno Brandsma
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  • Thank you Henno. I made the edit to the solution for the second part, as you indicated the proof for it. However, is the final paragraph needed for (ii)? – rudinsimons12 Mar 30 '20 at 15:19