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.
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?