The preimage of continuous function on a closed set is closed. My proof is very different from my reference, hence I am wondering is I got this right?
Apparently, $F$ is continuous, and the identity matrix is closed. Now we want to show that the preimage of continuous function on closed set is closed.
Let $D$ be a closed set, Consider a sequence $x_n \to x_0$ in which $x_n \in f^{-1}(D)$, and we will show that $x_0 \in f^{-1}(D)$.
Since $f$ is continuous, we have a convergent sequence
$$\lim_{n\to \infty} f(x_n) = f(x_0) = y.$$
But we know $y$ is in the range, hence, $x_0$ is in the domain. So the preimage is also closed since it contains all the limit points.
Thank you.
 A: Yes, it looks right.   Alternatively, given a continuous map $f: X \to Y$, if $D \subseteq Y$ is closed, then $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$ is open, so $f^{-1}(D)$ is closed.
A: Yes, your proof is correct, but since you are using sequences this works on metric spaces, not on topological ones.
A: Let me add an analogous approach that works for topological spaces. We need to introduce an additional concept that is basically a generalisation of sequences for topological spaces. 
Let $S$ be a directed set (reflexive, transitive and any pair of elements has a upper bound) and let $X$ be a topological space. Then a net is a function $x: S \rightarrow X$. We often use the notation $x_{\alpha}$ instead of $x(\alpha)$ for evaluation to mirror the notation used for sequences. And we say that a net converges to a point $x \in X$ when for all neighbourhoods of $x$ $N_{x}$ there exists an $\alpha_{0}$ such that
$$\left \{ x_{\alpha}:\alpha \geq \alpha_{0} \right \} \subset N_{x}.$$
We denote this fact by $x_{\alpha} \rightarrow x$.
Using this we can characterise closed set for general topological spaces in a "sequential manner". Let's compare the two characterisations that may be used:


*

*In a first countable space (every metric space is first countable), $A$ is closed iff for every sequence $(x_{n}) \subset A:x_{n} \rightarrow x$ we have $x \in A$.

*In a topological space (no additional separation assumptions are needed), $A$ is closed iff for every net $(x_{\alpha}) \subset A:x_{\alpha} \rightarrow x$ we have $x \in A$.


Note that we also have a theorem for nets that states that we can interchange a continuous function and the limiting operation. Thus allowing us to do exactly (notation wise) what 1LiterTears did in the initial post (consider the same setup, just replace sequence with net):
$$\text{lim}f(x_{\alpha}) = f(\text{lim}x_{\alpha})=f(x).$$
This of course allows us to also get the "same result" if we replace sequence by net in the initial statement.
In short, nets are a great way to move from first countable spaces to general topological spaces while still being able to recycle some of the "standard arguments" from analysis.
