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A function $f :\Bbb R\to\Bbb R$ is continuous on $\Bbb R$ if and only if $f^{-1}(F)$ is closed in $\Bbb R$ whenever $F$ is closed in $\Bbb R.$

My attempt so far:

We know that,

A function $f :\Bbb R\to\Bbb R$ is continuous on $\Bbb R$ if and only if $f^{-1}(G)$ is open in $\Bbb R$ whenever $G$ is open in $\Bbb R.$

Let $F$ be a closed set in $\Bbb R.$ This means that $F^c$ is open. Now, using the theorem above, we have, $f^{-1}(F^c)$ is open. Hence, $(f^{-1}(F^c))^c=f^{-1}(F)$ is closed.

Conversely, if for any closed set $F$, if we have, $f^{-1}(F)$ closed then let $F$ be a closed set. This means $F^c$ is open and as $f^{-1}(F)$ is closed so, $(f^{-1}(F))^c=f^{-1}(F^c)$ is also open. We thus have, $f^{-1}(F^c)$ is open whenever $F^c$ is open. But $F$ is arbitrary so is, $F^c.$ Hence, $f$ is continuous on $\Bbb R.$


Is the above solution correct? I think the part where I asserted, $(f^{-1}(F))^c=f^{-1}(F^c)$ needs a justification. I really can't come up with any. Is this assertion at all valid?

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About the double inclusion: If $x\in (f^{-1}(F^c))^c$, then $f(x)\notin F^c$, thus $f(x)\in F$, which means that $x\in f^{-1}(F)$. If you reverse the reasoning, you get the desired identiny.

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