Difference between revisions of "Compact"

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=Compact (Topology)=
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A topological space $$X$$ is '''compact'''' if every open covering of $$X$$ contains a finite subcollection that also covers $$X$$.
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Compactness is a homeomorphism invariant.
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If $$X$$ is a compact space and $$X/∼$$ is a quotient space, then $$X/∼$$ is compact.
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If $$A$$ is a subspace of a compact space $$X$$ and $$A$$ is a closed subset in $$X$$, then $$A$$ is compact.
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(EVT = Extreme Value Theorem) Let $$X$$ be a compact space, and let $$f: X → (ℝ,𝒯_{\rm Eucl})$$ be a continuous function. Then there exist $$c,d ∈ X$$ such that for all $$p ∈ X, f(c) ≤ f(p) ≤ f(d)$$.
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A continuous image of a compact space is compact. That is, if $$X,Y$$ are topological spaces, and if $$X$$ is compact and $$f: X → Y$$ is a continuous surjective function, then $$Y$$ is compact.
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Let $$A$$ be a subspace of a topological space $$(X,𝒯_X)$$. The space $$A$$ is compact if and only if for every collection $$𝒞$$ of open sets in $$X$$ satisfying $$A ⊆ ∪_{C ∈ 𝒞} C$$, there is a finite subcollection $$𝒟 ⊆ 𝒞$$ such that $$A ⊆ ∪_{D ∈ 𝒟} D$$.
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[[Category:Point-Set Topology]]

Latest revision as of 22:50, 21 May 2023

Compact (Topology)

A topological space $$X$$ is compact' if every open covering of $$X$$ contains a finite subcollection that also covers $$X$$.

Compactness is a homeomorphism invariant.

If $$X$$ is a compact space and $$X/∼$$ is a quotient space, then $$X/∼$$ is compact.

If $$A$$ is a subspace of a compact space $$X$$ and $$A$$ is a closed subset in $$X$$, then $$A$$ is compact.

(EVT = Extreme Value Theorem) Let $$X$$ be a compact space, and let $$f: X → (ℝ,𝒯_{\rm Eucl})$$ be a continuous function. Then there exist $$c,d ∈ X$$ such that for all $$p ∈ X, f(c) ≤ f(p) ≤ f(d)$$.

A continuous image of a compact space is compact. That is, if $$X,Y$$ are topological spaces, and if $$X$$ is compact and $$f: X → Y$$ is a continuous surjective function, then $$Y$$ is compact.


Let $$A$$ be a subspace of a topological space $$(X,𝒯_X)$$. The space $$A$$ is compact if and only if for every collection $$𝒞$$ of open sets in $$X$$ satisfying $$A ⊆ ∪_{C ∈ 𝒞} C$$, there is a finite subcollection $$𝒟 ⊆ 𝒞$$ such that $$A ⊆ ∪_{D ∈ 𝒟} D$$.