# Does there exist a class of real-valued upper semicontinuos functions on $X$ such that $\mathcal{F}$ is countable?

Ian Morris quoted the following:

For any upper semi-continuous function $f \colon X \to [-\infty,+\infty)$ defined on a nonempty topological space $X$ there exists a nonempty set $\mathcal{F}\subset C(X,\mathbb{R})$ such that $f(x)=\inf\{g(x)\colon g \in \mathcal{F}\}$ for every $x \in X$. If $X$ is metrisable then $\mathcal{F}$ may be taken to be countable.

I am interested in the bold sentence: If $X$ is metrizable, then the class of continuous functions $\mathcal{F}$ can be countable.

Question: If we define a function $f:X \rightarrow [\infty,+\infty)$ such that for any real number $c \in \mathbb{R},$ its pre-image $f^{-1}(-\infty,c)$ is a $F_{\sigma}$ set, can we get a class $\mathcal{F}$ of real-valued upper semicontinuos functions on $X$ such that $f(x) = \inf \{ g(x):g \in \mathcal{F} \}$ and $\mathcal{F}$ is countable?

• Consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 0$ for $x \leq 0$ and $f(x) = 1$ for $x > 0$. Is it the infimum of a class of upper semicontinuous functions? – Nik Weaver Apr 13 '17 at 4:33
• @NikWeaver: The $f$ is not the infimum of a class of upper semicontinuous functions, as the infimum should be upper semicontinuous. – Idonknow Apr 13 '17 at 12:04

Based on the comment by Nik Weaver, the answer to my question is negative, that is, there exists a function $f:X \rightarrow \mathbb{R}$ such that it is not an infimum of any upper-semicontinuous functions:
Define $f:X \rightarrow \mathbb{R}$ such that $f(x) = 0$ if $x \leq 0$ and $f(x) =1$ if $x>0.$ Clearly $f$ is lower semicontinuous everywhere and not continuous at $x=0.$
If $f$ is the infimum of a class of upper semicontinuous functions, then $f$ should be upper semicontinuous. A lower and upper semicontinuous function should be continuous, a contradiction. So $f$ is not the infimum of a class of upper semicontinuous functions.