Difference between revisions of "Compact"
Kfagerstrom (talk | contribs) |
Kfagerstrom (talk | contribs) |
||
| Line 1: | Line 1: | ||
=Compact (Topology)= | =Compact (Topology)= | ||
| − | |||
| − | 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. | + | 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$$. | ||
Revision as of 22:22, 21 May 2023
Compact (Topology)
A topological space $$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 $$A ⊆ ∪_{D ∈ 𝒟} D$$.
