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Compact Sets in the Order Topology contain their Extrema

Last updated Nov 6, 2022

# Statement

Let (X,)(X, \leq) be a Total Ordering

Total Ordering

Definition A (T,)(T, \leq) is a if x,yT\forall x,y \in T, either xyx \leq y or yxy \leq x....

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and let KXK \subset X be Compact

Compact

Definition Let XX be a . We say KXK \subset X is if every of KK can be reduced...

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. Then supK\sup\limits K and infK\inf\limits K both exist. Furthermore, supKK\sup\limits K \in K and infKK\inf\limits K \in K.

# Proof

Recall that A Nonempty Set is Compact in the Order Topology iff it is Tightly Bounded and Complete

. Therefore, KK has an Upper Bound

Upper Bound

Definition Let (P,)(P, \leq) be a and let APA \subset P. xPx \in P is an for AA if...

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and a Lower Bound

Lower Bound

Definition Let (P,)(P, \leq) be a and let APA \subset P. xPx \in P is an for AA if...

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. Because the Order Topology is Hausdorff

Order Topology is Hausdorff

Statement Let (X,)(X, \leq) be a with the . Then XX is . Proof We break into cases on XX. X=1|X| = 1:...

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and Compact Sets in Hausdorff Spaces are Closed

Compact Sets in Hausdorff Spaces are Closed

Statement Let XX be a . Then if KXK \subset X is , KK is . Proof Let xKCx \in K^{C}. Because...

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, we know KK is Closed

Closed

Definition Suppose (X,τ)(X, \tau) is a . Then KXK \subset X is if KCK^{C} is ....

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.