The set of Upper semi-continuous functions as a ring.

Question

I should recall that the surgenfery topology on the real numbers is denoted by $\mathbb{R}_l$, and has the set
{$[a , b): a,b \in \mathbb{R} $} as it's base.

If $X$ is a topological space, an upper semi-continuous real function on $X$ can be interpret as a continuous function from $X$ into $\mathbb{R}_l$.

The set of all upper semi-continuous real functions on $X$ is denoted by $USC(X)$.

We could easily see that if we consider $X=\mathbb{R}$ then the set $USC(\mathbb{R})$ is not a ring with pointwise addition and multiplication. because there is $f \in USC(\mathbb{R}) $ so that $-f \notin USC(\mathbb{R}) $. Indeed it is not a Group.

Question.For what condition(s) on $X$, the set $USC(X)$ constructs a ring structure we the pointwise addition and multiplication?

PS:I am looking for the topological condition(s) $P$ on $X$ so that, $USC(X)$ is a ring iff $X$ has the property $P$.

Thank you so much for noticing to my Question.

Answer 1

Any set $U$ which is open and not closed provides an example: the characteristic function $\chi_U$ is in $USC(X)$, but $-\chi_U$ is not, or perhaps the other way around.

Topologies where every open set is closed, and vice versa, come from partitions of the base set, with open and closed sets being arbitrary unions of parts. Thus, a semicontinuous function from such a topological space to $\mathbb R$ is just a function $X \to \mathbb R$ that is constant on the parts, or a function from the set of parts to $\mathbb R$.

Since the space of functions from any set to $\mathbb R$ form a ring under pointwise addition and multiplication, $USC(X)$ is a ring if and only if the topology on $X$ is of this type.