# Preimage under $f$ of any closed set is closed.

Let $$f: S \to \mathbb{R}$$, where $$S \subset \mathbb{R}$$. Prove that the preimage under $$f$$ of any open set is open $$\implies$$ preimage under $$f$$ of any closed set is closed. My approach: Take $$U \subset \mathbb{R}$$ closed (ie $$U = [a, b]$$). Then $$U = (a, b) \cup$${$$a, b$$}$$\implies$$ By assumption $$f^{-1}(U \setminus$${$$a, b$$}$$)$$ is open in $$S$$ since $$U \setminus$${$$a, b$$} is open $$\implies$$ $$f^{-1}(U) \setminus f^{-1}($${$$a, b$$}$$)$$ is open in $$S$$ $$\implies$$ $$S \setminus \big(f^{-1}(U) \setminus f^{-1}($${$$a, b$$}$$) \big )$$ is closed in $$S$$ $$\implies \big(S \setminus (f^{-1}(U)\big) \cap \big (S\setminus f^{-1}($${$$a, b$$}$$)\big)$$ is closed in $$S$$.

I need to show that $$S \setminus f^{-1}(U)$$ is open in S. However I am not sure how to proceed after the last step above. Any help would greatly be appreciated.

• " Take U⊂R closed (ie U=[a,b])" i.e. means "in other words" not "for example" a closed set need not be a closed interval. Nov 7, 2018 at 16:42

You are assuming that the only closed subsets of $$\mathbb R$$ are the intervals of the form $$[a,b]$$. That is wrong.
If $$U\subset\mathbb R$$ is closed, then $$\mathbb{R}\setminus U$$ is open. Furthermore, $$f^{-1}(\mathbb{R}\setminus U)=S\setminus f^{-1}(U)$$. Therefore, since $$f^{-1}(\mathbb{R}\setminus U)$$ is open, $$f^{-1}(U)$$ is closed.
• I see. Is there an example of a closed set in $\mathbb{R}$ not of the form $[a, b]$ so that I can visualize it? Nov 7, 2018 at 16:47
• $\{0,1\}$, for instance. Nov 7, 2018 at 16:48
• $(a,b)$ is open so $(a,b)^c = (-\infty,a]\cup [b,\infty)$ is another. But so is a set of point $\{a,b,c,d\}$ or $\mathbb Z$. or combos thereof. The set $\{\frac 1n|n\in \mathbb N\}\cup\{0\}$ is an interesting one. It's closed because in contains it's limit point $0$. But its also closed because $(-\infty,0)\cup [\cup_{n\in \mathbb N}(\frac 1{n+1},\frac 1n)]$ is open. Nov 7, 2018 at 17:00
• "$f^{-1}(\mathbb{R}\setminus U)=S\setminus f^{-1}(U)$" Perhaps the OP should be asked to prove this? Nov 7, 2018 at 17:02