General Topology/Order topology and semicontinuity

Definition (order topology):

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

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

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

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

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

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

Proof: We have

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

Theorem (Weierstraß-type theorem):

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

Proof: The sets

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

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

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

But

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

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