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

compactnessgeneral-topologymetric-spacesproduct-space

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

Best Answer

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.