Projection of a closed set in $\mathbb{R}^2$ to one of the factors need not be closed.

Question

$\def\R{{\mathbb R}}$ 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.

Answer 1

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).

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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.

Answer 2

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.