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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.

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    $\begingroup$ " 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. $\endgroup$
    – fleablood
    Nov 7, 2018 at 16:42

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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.

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  • $\begingroup$ 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? $\endgroup$ Nov 7, 2018 at 16:47
  • $\begingroup$ $\{0,1\}$, for instance. $\endgroup$ Nov 7, 2018 at 16:48
  • $\begingroup$ $(a,b)$ 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. $\endgroup$
    – fleablood
    Nov 7, 2018 at 17:00
  • $\begingroup$ "$f^{-1}(\mathbb{R}\setminus U)=S\setminus f^{-1}(U)$" Perhaps the OP should be asked to prove this? $\endgroup$
    – fleablood
    Nov 7, 2018 at 17:02

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