# 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.

• Your last sentence needs to be rephrased. $y$ must be in $D$ (since $D$ is closed and each $f(x_n)\in D$); thus $x_0$ is in $f^{-1}(D)$. Jul 7, 2013 at 1:33
• Thanks @DavidMitra - that indeed is what I intended. Jul 7, 2013 at 1:35
• would you add the reference for your reference? Sep 11, 2018 at 23:09

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.

• However, why you can make the assertion that $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$? Jul 7, 2013 at 2:01
• @Jellyfish Just do an element chase to prove that they're equal. It has nothing to do with topology.
– Ink
Jul 7, 2013 at 2:09
• $X=f^{-1}(Y)$ requires $f$ to be a total function. Apr 1, 2018 at 7:42
• Would you please add some reference on this? Sep 11, 2018 at 23:08
• That doesn't seem necessary, as generally the definition of a function includes totality. Mar 17, 2020 at 17:16

Yes, your proof is correct, but since you are using sequences this works on metric spaces, not on topological ones.

• This answer interests me. If I want to work on topological ones, how should I do? Does the discussion on metric spaces not be included in the topological space? Aug 4, 2016 at 20:50
• @sleevechen I think you probably need the topological space to be Hausdorff Mar 23, 2020 at 18:40

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.