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 $\Bbb R.$

Question

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?

Does this answer your question? Show that for any subset $C\subseteq Y$, one has $f^{-1}(Y\setminus C) = X \setminus f^{-1}(C)$ — Anne Bauval, Nov 09 '23 at 16:51
If $f:X\to Y$, then $X\setminus f^{-1}(B)=f^{-1}(Y\setminus B)$ for any $B\subseteq Y$. This is a general property of functions, and you should try to prove it yourself — Lorago, Nov 09 '23 at 16:52

score 0 · Answer 1 · answered Nov 09 '23 at 17:07

0

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.

answered Nov 09 '23 at 17:07

SK_

575

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 $\Bbb R.$

1 Answers1