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.

  • $\begingroup$ 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 $f^{-1}(D)$. $\endgroup$ – David Mitra Jul 7 '13 at 1:33
  • $\begingroup$ Thanks @DavidMitra - that indeed is what I intended. $\endgroup$ – 1LiterTears Jul 7 '13 at 1:35
  • $\begingroup$ would you add the reference for your reference? $\endgroup$ – 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.

  • $\begingroup$ Thank you for presenting me a third way.. In fact this is my original thought, but I was not able to went through. Again, thanks a lot! $\endgroup$ – 1LiterTears Jul 7 '13 at 1:58
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    $\begingroup$ However, why you can make the assertion that $X \setminus f^{-1}(D) = f^{-1}(Y \setminus D)$? $\endgroup$ – 1LiterTears Jul 7 '13 at 2:01
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    $\begingroup$ @Jellyfish Just do an element chase to prove that they're equal. It has nothing to do with topology. $\endgroup$ – Ink Jul 7 '13 at 2:09
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    $\begingroup$ $X=f^{-1}(Y)$ requires $f$ to be a total function. $\endgroup$ – Witiko Apr 1 '18 at 7:42
  • $\begingroup$ Would you please add some reference on this? $\endgroup$ – Albert Chen Sep 11 '18 at 23:08

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

  • $\begingroup$ 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? $\endgroup$ – sleeve chen Aug 4 '16 at 20:50

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