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My question arises from the combination of both following theorems:

Theorem 1: Every open set is a continuous image of a closed set. That is, for every open set $A\subseteq \mathbb{R}^m$ there is a closed set $C\subseteq \mathbb{R}^n$ and a continuous function $f\colon \mathbb{R}^n \to \mathbb{R}^m$ such that $f\left( C \right) = A$

Theorem 2: A function is continuous iff it's preimage on every open set is open. That is, $f\colon \mathbb{R}^n \to \mathbb{R}^m$ is continious iff $f^{-1}\left( A \right)$ is open for every open set $A\subseteq \mathbb{R}^m$

Let $A\subseteq \mathbb{R}^m$ be an open set. From Theorem 1 we know that there is a closed set $C\subseteq \mathbb{R}^n$ and a continuous function $f\colon \mathbb{R}^n \to \mathbb{R}^m$ such that $f\left( C \right) = A$ .

We now have a continuous function $f$ and an open set $A\subseteq \mathbb{R}^m$ , so by Theorem 2 it's preimage $f^{-1}\left( A \right)$ has to be open. But $f^{-1}\left( A \right)=C$ is a closed set.

What am I missing here?

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Jon
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1 Answers1

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Just because $f(C)=A$ does not mean that $f^{-1}(A) = C$. You only conclude that $f^{-1}(A) \supset C$, but the preimage could be larger.

Ted
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  • So take $f\left|_{C}\right.$ – Jon Nov 07 '18 at 07:21
  • But then $C$ is both open and closed in $C$, so there's no contradiction. – Ted Nov 07 '18 at 07:21
  • What do you mean by both open and closed in $C$? – Jon Nov 07 '18 at 07:22
  • And how you conclude that $C \subset f^{-1}\left( A \right)$? – Jon Nov 07 '18 at 07:26
  • If you restrict the function to $C$, then you have to restrict your open and closed sets to their intersection with $C$ (the subspace topology) for theorems 1 and 2 to remain valid. A set is "open in $C$" if it is the intersection of $C$ with an open set in $\mathbb{R}^n$ (similar definition for "closed in $C$"). – Ted Nov 07 '18 at 07:26
  • That's just the definition of $f^{-1}$ ... – Ted Nov 07 '18 at 07:27
  • I mean how you conclude that $C\subsetneq f^{-1}\left(A\right)$? – Jon Nov 07 '18 at 07:29
  • First, note that just because you conclude that $C$ is both open and closed, that is not immediately a contradiction (https://en.wikipedia.org/wiki/Clopen_set). But if $C$ is not clopen, then you do get a contradiction and therefore we must have $C \subsetneq f^{-1}(A)$. In $\mathbb{R}^n$, the only clopen sets are the empty set and the entirety of $\mathbb{R}^n$. – Ted Nov 07 '18 at 07:31