I've tried looking for this question on this site, but I can't seem to find it. But if anyone can direct me to it, that would be great. But I'll pose my question in the mean time.
As the title states, if $f^{-1}(\alpha, \infty)$ is open, then show that $f$ is lower-semicontinuous.
Starting with the basics. The definition according to my professor gave on lower-semicontinuous is the following:
Given a function $f: X \to \mathbb{R}$ on a topological space $X$, $f$ is lower-semicontinuous if, for any $x \in X$ and for any $\epsilon > 0$, there is a neighborhood $N$ of $x$ such that
$f(x) - \epsilon < f(x')$ for all $x' \in N$.
I proved the conversed of this statement, but for this direction I seem to be stuck, been thinking for it for an hour or two. To my understanding a neighborhood of a point $x$ is a subset of $X$ such that it contains an open set which has $x$.
To me, it seems that I must consider if $x \in f^{-1}(\alpha, \infty)$ or $x \in f^{-1}(-\infty, \alpha)$. I started with the consideration of $x \in f^{-1}(\alpha, \infty)$. So by definition of pre-image, $f(x) \in (\alpha, \infty)$. We have that $\alpha < f(x) < \infty$. So it seems to me that I want a neighborhood of $x$ such that it satisfies $f(x) - f(x') < \epsilon$ for all $x' \in$ nieghborhood of $x$. Fix $x \in X$. Then I went on the path of suppose that $x \in f^{-1}(\alpha, \infty)$ and using that using $f^{-1}(\alpha, b)$ where $b = f(x)$. Since $f^{-1}(\alpha, \infty)$ can be written as the union (index starting at $k$) of $f^{-1}(\alpha, k)$ for $k > \alpha$ and $k \in \mathbb{R}$, then $f^{-1}(\alpha, b)$ must be open. Then the problem is that I can't get a neighborhood of $x$ such that $f(x) - f(x') < \epsilon$. I get a neighborhood of $x$ but it has elements such that $f(x) - f(x') \not<\epsilon$. I'll currently will update the progress of my work. Hopefully I can progress somewhere. Thanks.
