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Possible Duplicate:
Show the iff statement of lower semicontinuous

Given $X\subseteq \Bbb R^m, f:X\to\Bbb R$ and $x\in X$, we say $f$ is lower semicontinous (l.s.c for short) at x if

$\forall \varepsilon>0\ \exists\ \delta >0\ \forall \in B(\delta,x), \ f(x)\le f(y)+\varepsilon$.

I wish to show:
If $X$ is closed, then $f$ is l.s.c if and only if the set $f^{-1}((-\infty,r]):=\{a\in X:f(a)\le r\}$ is closed for each $r\in \Bbb R$.

I have already proved that:
If $X$ is closed, then $f$ is l.s.c if and only if the set $f^{-1}((r,+\infty):=\{a\in X:f(a)\gt r\}$ is open for each $r\in \Bbb R$.

I wonder how could I use this to prove that statement I wish to prove, which is in the closed sets. The following was what I tried:

For the forward: since we have showed that if f is l.s.c, then the set $f^{-1}((r,\infty)):=\{a\in X:f(a)\gt r\}$ is open for each $r\in \Bbb R$. Then, since the complement of a open set is closed, then, we have if f is l.s.c, then $(f^{-1}((r,+\infty))^{C}$ is closed. But $(f^{-1}((r,+\infty)))^{C}=f^{-1}((r,+\infty)^{C})=f^{-1}((-\infty,r])$. Thus this completes the forward proof of the statement.
I don't know whether this is right or not.

And, I have no idea how to show the backward based on the open sets since I just feel it have the same thing as what I prove above.

Hope for any help, please. Thank you!

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