Upper semicontinuous functions to $\mathbb{N}$ are locally constant on a dense subset

Question

Let us take $f : X \rightarrow \mathbb{N}$ an upper semicontinuous function. In Wikipedia - Semi-continuity it is said that such a function must be locally constant on a dense open subset. I don't know how to prove it. I tried playing with $\mathbb{R}_{usc}$, where the subscript means upper semicontinuity, i.e. the open subsets are of the form $(- \infty, x)$ , $x \in \mathbb{R}$ and considering $i \circ f$, where $i : \mathbb{N} \hookrightarrow \mathbb{R}_{usc}$ is the inclusion with respect to the subspace topology, but it didn't work. Any help?

Answer 1

I stumble myself again this very same question after reading the Wikipedia page independently, so I decided to give it a try.

Let $S = \{x \in X : f \textrm{ is locally constant at } x\}.$ Suppose $S$ is not dense on X. Then, $\bar{S}^c$ is a nonempty open set such that $f$ is not locally constant at any of its points. Define $$n_0 := \min_{x \in \bar{S}^c} f(x). $$ This number is well defined because the range of $f$ is a subset of $\Bbb N.$ In addition, consider the set

$$V:= \{x \in X: f(x) < n_0 + \frac{1}{2}\}.$$ Because $f$ is u.s.c, the set $V$ is open. Finally, define $U:= V \cap \bar{S}^c.$ Then, $U$ is nonempty, open, and $U \subseteq \bar{S}^c.$ Furthermore, by the definition of $n_0$ and $U,$ we must have $f(x) = n_0$ for every $x\in U.$ But this contradicts the fact that $f$ is not locally constant on $\bar{S}^c.$ The statement follows.