Projection of $M \times N \to N$ is closed where $M$ is compact

Question

I'm trying to prove that if $M$ is compact and $F \subset M \times N$ is closed, then the projection $\pi: M \times N \to N$ is a closed map (if $F \subset M \times N$ is closed, then $\pi(F)$ is closed). I don't know about the tube lemma, so I'm trying to use sequences.

If $(x_n, y_n)_{n \in \mathbb{N}} \subset F$, then exists $(x,y) \in F$ such that $(x_n, y_n)$ converges to $(x,y)$ because $F$ is closed. So what I'm trying to show is that $(\pi(x_n, y_n))_{n \in \mathbb{N}} \subset \pi(F)$ converges to some point in $\pi(F)$ but I'm kinda stuck, I have no idea where compactness comes in place here. I'm probably missing something really obvious. Any tips? Thanks in advance!

Answer 1

Since you are using sequences I would assume that you are working in metric spaces. (Sequences are not adequate in general toploogical spaces. You would have to use nets but the argument is not very different).

You have to start with $(x_n,y_n) \in F$ with $\pi (x_n,y_n)$ converging to some $z \in N$. This gives $y_n \to z$. There is a subsequence $(x_{n_k})$ converging to some $x \in M$ (because $M$ is compact). Now $z=\lim \pi (x_{n_k},y_{n_k})=\pi (x,z) $ since $\pi$ is continuous. Note that $(x,z) \in F$ because $F$ is closed. Hence, $z \in \pi (F)$ and we are done.