View source for General Topology/Order topology and semicontinuity

{{definition|order topology|Let <math>(S, \le)</math> be poset. The '''order topology''' on <math>S</math> is the topology which has as a subbasis the sets
:<math>(a, b) := \{x \in S| a < x < b\}</math>.}}

{{proposition|half-open intervals on lattices form a topology base|Let <math>(S, \le)</math> be a lattice. Then the sets
:<math>(a, b) := \{x \in S| a < x < b\}</math>
form a <math>\pi</math>-system; in particular, they form a topology base.}}

{{proof|We have
:<math>(a, b) \cap (c, d) = (a \vee c, b \wedge d)</math>.}}

{{theorem|Weierstraß-type theorem|Let <math>X</math> be a compact topological space, and let <math>(S, \le)</math> be a lattice. Let <math>f: X \to S</math> be continuous with respect to the order topology on <math>S</math>. Then <math>f(X)</math> is bounded in <math>S</math>.}}

{{proof|The sets
:<math>f^{-1}((a,b))</math>
form an open cover of <math>X</math>, where <math>a, b</math> range over <math>S</math>. By compactness, we may find a finite subcover
:<math>X = f^{-1}((a_1, b_1)) \cup \cdots \cup f^{-1}((a_n, b_n))</math>.
But
:<math>(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)</math>,
so that <math>f</math> maps every point in <math>X</math> into the latter interval.}}

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