# For every closed set $C$, $f^{-1}(C)$ is closed. Is $f$ necessarily continuous?

I am trying to prove this using the definition of continuity (and complements) where $f$ is continuous iff for every open set $U$, $f^{-1}(U)$ is also open. But I am uncertain about my proof when I try to write it formally. If somebody can present a formal proof, it will be really appreciated.

Thanks!

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A function is continuous if and only if the pre-image of an open set is open. Let $f:X\to Y$ be such that for every closed set $C$ in $Y$ the set $f^{-1}(C)$ is closed. Now, consider an open set $U$. Then $Y-U$ is closed by definition, so $f^{-1}(Y-U)$ is closed. But $$f^{-1}(Y-U)=f^{-1}(Y)-f^{-1}(U)=X-f^{-1}(U).$$ Since $X-f^{-1}(U)=(f^{-1}(U))^c$ is closed, its complement, i.e., $f^{-1}(U)$ is open, thus $f$ is continuous.

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it looks good at first sight but when you consider U as an open set and then conclude from there that Y-U is closed by definition (what definition are you talking about). If Y is the set of all reals and U = (0,infinity), then i think Y-U is not necessarily closed. what are your thoughts? – UH1 Dec 7 '12 at 21:02
Closed means that it contains all of its limit points (where a limit point is defined to be a point such that any open set containing it contains a point distinct from it in the bigger set), so in fact $(-\infty,0]$ is closed, since for any point in $(-\infty,0]$, when we look at any open set containing it (even $0$), there are points in $(-\infty,0]$ that aren't $x$. – Clayton Dec 7 '12 at 21:03
And one more thing, are you sure this equality holds for any function $f^{-1}(Y-U)=f^{-1}(Y)-f^{-1}(U)$ Can we just quote this? – UH1 Dec 7 '12 at 21:12
Yes, it is fairly easy to prove even. I would guess it's fair game just to quote it. – Clayton Dec 7 '12 at 21:14
do you mind writing a quick sketch just so we can "close" this matter? i will be very grateful. thanks! – UH1 Dec 7 '12 at 21:16

The two key facts are: (a) a set is open iff its complement is closed; and (b) inverse image preserves complements, i.e. if $f:X\to Y$, and $U \subseteq Y$, then $f^{-1}(U^c) = (f^{-1}(U))^c$.

Given this: suppose $U \subseteq Y$ is open. We need to show that $f^{-1}(U)$ is open; equivalently, that its complement is closed. But its complement is $(f^{-1}(U))^c = f^{-1}(U^c)$, the inverse image of a closed set, so is closed by the assumption on $f$.

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Suppose that $f:X\to Y$ is such that $f^{-1}[F]$ is closed in $X$ whenever $F$ is closed in $Y$. Let $U$ be an arbitrary open set in $Y$. Then $Y\setminus U$ is closed in $Y$, so $f^{-1}[Y\setminus U]$ is closed in $X$, and therefore $X\setminus f^{-1}[Y\setminus U]$ is open in $X$. Now show that $X\setminus f^{-1}[Y\setminus U]=f^{-1}[U]$, and you’re done.

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Do you know how to prove the last equality you quote? because I am having the same problem with @clayton's solution below – UH1 Dec 7 '12 at 21:14
@UH1: Just chase elements: $x\in f^{-1}[U]$ iff $f(x)\in U$ iff $f(x)\notin Y\setminus U$ iff $x\notin f^{-1}[Y\setminus U]$ iff $x\in X\setminus f^{-1}[Y\setminus U]$. – Brian M. Scott Dec 7 '12 at 21:23
thanks, got it! – UH1 Dec 7 '12 at 21:27
@UH1: Excellent! – Brian M. Scott Dec 7 '12 at 21:33

Suppose $f:X \to Y$ satisfies the condition, and let $U \subset Y$ be an open set. Then $U^c$ is closed subset of $Y$, and so we have that $f^{-1}(U^c)$ is closed in $X$. But $f^{-1}(U^c) = f^{-1}(U)^c$, and so $f^{-1}(U)$ is open since its complement is closed.

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