As part of a bigger problem the following argument is applied:
Let $X$ be a regular topological space I is a lower semicontinuous function $X \rightarrow [0,+\infty]$. (...)
Let $x \in X$ be fixed and let $c<I(x)$. By definition of lower semicontinuity the set $C=\{I \le c\}$ is closed One can separate $x$ from $C=\{I \le c\}$. Thus there exists an open set $G$ containing $x$ such that $\overline{G}\subset \{I >c \}$.
I am having trouble with the last sentence. As far as I know, as stated in my notes:a topological space is regular if points and closed sets can be separated by disjoint open neighborhoods. Therefore I would say that there exists disjoint open sets $G$ and $H$, such that $x\in G$, $C \subseteq H$. But in the problem I am reading they seem to use some property or equivalence. It looks like that regularity implies or is equivalent to :
"for every $x \in X$ and closed set $C \subseteq X$ $\exists$ open $G$ with $x\in G$ s.t $\overline{G}\subset C^C$"
Is this an alternative definition or property ? Or just an implication? I tried proving it but I am not getting anything meaningful. Can someone help me prove it?
A well-known alternative characterization of regularity is this:
For each $x \in X$ and each open $U \subset X$ such that $x \in U$ there exists an open $V \subset X$ such that $x \in V$ and $\overline V \subset U$.
This is easy to prove and can be found in any textbook on general topology.
It is precisely that what you want to use.