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 $f^{-1}(D)$. – David Mitra Jul 7 '13 at 1:33
• Thanks @DavidMitra - that indeed is what I intended. – 1LiterTears Jul 7 '13 at 1:35
• would you add the reference for your reference? – Albert Chen Sep 11 '18 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 $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$? – 1LiterTears Jul 7 '13 at 2:01
• $X=f^{-1}(Y)$ requires $f$ to be a total function. – Witiko Apr 1 '18 at 7:42