Upper semicontinuity

This is a notion that I had to look up whenever I face it. Hence, I decided to write it down in my own terms so that I don't have to look it up as many times. In Vakil (11.4), this notion is defined as follows: given a topological space $X,$ a function $f : X \rightarrow \mathbb{R}$ is called upper semi-continuous if $f^{-1}((-\infty, a)) \subset X$ is open for any $a \in \mathbb{R}.$ This is a formal definition, and experts don't seem to think this way. This is a condition on $f$ so that it can only jump up when taking limits.

Theorem. The following are equivalent:

• $f$ is upper semi-continuous;
• for any sequence $(x_{n})$ in $X$ (with possibly uncountable indices $n$) such that $f(x_{n}) \geq f(x),$ if $x_{n} \rightarrow x$ in $X,$ then $f(x_{n}) \rightarrow f(x)$ in $\mathbb{R}.$

Proof. Let $f$ be upper semi-continuous. Given a sequence $(x_{n})$ in $X$ such that $f(x_{n}) \geq f(x),$ if $x_{n} \rightarrow x$ in $X,$ fix any $\epsilon > 0.$ Then $f^{-1}((-\infty, f(x) + \epsilon)) \subset X$ is open by upper semi-continuity of $f$ and since it contains $x,$ knowing that $x_{n} \rightarrow x$ in $X,$ we may find some $x_{n} \in f^{-1}((-\infty, f(x) + \epsilon)).$ This implies that $$f(x_{n}) \in (-\infty, f(x) + \epsilon) \subset \mathbb{R}.$$ Since $f(x_{n}) \geq f(x) > f(x) - \epsilon,$ we have $$f(x_{n}) \in (f(x) - \epsilon, f(x) + \epsilon) \subset \mathbb{R}.$$ This shows that $f(x_{n}) \rightarrow f(x)$ in $\mathbb{R}.$

Conversely, suppose the second condition above. Fix $a \in \mathbb{R},$ and we aim to show that $f^{-1}((-\infty, a)) \subset X$ is open. Fix any $x \in f^{-1}((-\infty, a)).$ We want to show that there is an open neighborhood $U \ni x$ in $X$ such that $U \subset f^{-1}((-\infty, a)).$ We now work by contradiction and say that there is no such $U.$ In other words, for any neighborhood $U_{n} \ni x$ in $X,$ there exists $x_{n} \in U_{n}$ such that $f(x_{n}) \geq a > f(x).$ However, using the second condition, this implies that $$f(x) = \lim_{n \rightarrow \infty}f(x_{n}) \geq a > f(x),$$ a contradiction. This finishes the proof. $\Box$