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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.

Martin Sleziak
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MathNewbie
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    What is your definition of "lower semicontinuous"? How one proves a function has the property depends on the definition you are using, and there are several alternative ways of defining it here. – Arturo Magidin Mar 22 '12 at 21:45
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    Presumably it's the same definition as in this question. As a bonus, the current question is an almost immediate corollary of that one. – Chris Eagle Mar 22 '12 at 22:01

2 Answers2

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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?

Brian M. Scott
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  • Sorry, I meant to put that $f: X \to \mathbb{R}$, where $X$ is a topological space. I'll edit that now. From Chris's comment, I was just wondering if we can argue directly (without using that $f^{-1}[(a, \infty)]$ is open) that $f$ is lower-semicontinuous given that $f^{-1}(-\infty, a])$, but I know that the complement makes life easier. – MathNewbie Mar 22 '12 at 22:21
  • @MathNewbie: Since your definition of lower semicontinuity is phrased in terms of open sets, while the condition that you’re assuming is one on closed sets, you will at some point have to pass to a complement. – Brian M. Scott Mar 22 '12 at 22:28
  • Ah thanks Brian. That was the response I was looking for. – MathNewbie Mar 22 '12 at 22:30
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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.

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