General Topology/Order topology and semicontinuity

Definition (order topology):

Let ${\displaystyle (S,\leq )}$ be poset. The order topology on ${\displaystyle S}$ is the topology which has as a subbasis the sets

${\displaystyle (a,b):=\{x\in S|a.

Proposition (half-open intervals on lattices form a topology base):

Let ${\displaystyle (S,\leq )}$ be a lattice. Then the sets

${\displaystyle (a,b):=\{x\in S|a

form a ${\displaystyle \pi }$-system; in particular, they form a topology base.

Proof: We have

${\displaystyle (a,b)\cap (c,d)=(a\vee c,b\wedge d)}$. ${\displaystyle \Box }$

Theorem (Weierstraß-type theorem):

Let ${\displaystyle X}$ be a compact topological space, and let ${\displaystyle (S,\leq )}$ be a lattice. Let ${\displaystyle f:X\to S}$ be continuous with respect to the order topology on ${\displaystyle S}$. Then ${\displaystyle f(X)}$ is bounded in ${\displaystyle S}$.

Proof: The sets

${\displaystyle f^{-1}((a,b))}$

form an open cover of ${\displaystyle X}$, where ${\displaystyle a,b}$ range over ${\displaystyle S}$. By compactness, we may find a finite subcover

${\displaystyle X=f^{-1}((a_{1},b_{1}))\cup \cdots \cup f^{-1}((a_{n},b_{n}))}$.

But

${\displaystyle (a_{1},b_{1})\cup \cdots \cup (a_{n},b_{n})\subseteq (a_{1}\wedge \cdots \wedge a_{n},b_{1}\vee \cdots \vee b_{n})}$,

so that ${\displaystyle f}$ maps every point in ${\displaystyle X}$ into the latter interval. ${\displaystyle \Box }$