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So I've been thinking about this for the last two hours, but I am stuck.

Suppose $f:X \to \mathbb{R}$ where $X$ is a topological space.

$f$ is said to be semicontinuous if for any $x \in X$ and $\epsilon > 0$, there is a neighborhood of $x$ such that f(x) - f(x') < \epsilon for all x' in the neighborhood of $x$.

The question is the if $f^{-1}((-\infty,a])$ is closed for $a \in \mathbb{R}$, then $f$ is lower semi-continuous.

I started with choosing an $x \in f^{-1}((-\infty,a])$ and letting $\epsilon > 0$. So far I don't know much characterization of closed sets in a topological space except it is the complements of open sets.

Not sure if this is correct, but I approached this problem with the idea of nets. Since $f^{-1}(-\infty,a]$ is closed, then for each $x \in f^{-1}(-\infty,a]$, there's a net $\{x_i\}_{i \in I}$ such that it converges to $x$ (not sure if I'm allowed to do that). Pick any neighborhood of $x$ denote by $N_x$ (which will contain terms from $f^{-1}(-\infty,a]$), which contains an open set which has $x$. Let $f(x) = b$. Then N_{x'}:=[N_x \backslash f^{-1}(-\infty, b)] \backslash [\mathrm{boundary \ of \ this \ set \ to \ the \ left}]. So this gives me an open set such that it contains $x$ and such that f(x) - f(x') < \epsilon for all x' \in N_{x'}. So $f$ must be semi-continuous.

Not sure if there is more to know about closed sets in a topological space, except its complement is open.

Any hint on how to think about this is appreciated.

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    Presumably it's the same definition as in [this question](http://math.stackexchange.com/questions/123379/show-that-if-f-1-alpha-infty-is-open-for-any-alpha-in-mathbbr). As a bonus, the current question is an almost immediate corollary of that one.2012-03-22

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I’m guessing from what you’ve written that your definition of lower semi-continuity is such that you want to start with an arbitrary $x_0\in X$ and $\epsilon>0$ and show that there is an open nbhd $U$ of $x_0$ such that $f(x)>f(x_0)-\epsilon$ for every $x\in U$. You know that for any $a\in\Bbb R$, $f^{-1}[(-\infty,a]]$ is closed, so its complement, which is $f^{-1}[(a,\infty)]$, must be open. Take $a=x_0-\epsilon$, and let $U=f^{-1}[(a,\infty)]$. Does this $U$ meet the requirements?

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    Ah thanks Brian. That was the response I was looking for.2012-03-22
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Here I showed that $f$ is lower semi-continuous whenever $f^{-1}((\alpha,\infty))$ is open for all real $\alpha$. Here we have that $f^{-1}((-\infty,\alpha])$ is closed, so $f^{-1}((\alpha,\infty))=f^{-1}((-\infty,\alpha]^c)=(f^{-1}((-\infty,\alpha]))^c$ is the complement of a closed set, and thus is open.