Difference between revisions of "The image of a compact set is compact"
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(Created page with "{{Stub page|grade=A|msg=Flesh out with references, proof is small but easy and can wait}} __TOC__ ==Statement== Let {{Top.|X|J}} and {{Top.|Y|K}} be topological spaces, le...") |
(Added proof) |
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{{Requires references|grade=A|I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?}} | {{Requires references|grade=A|I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?}} | ||
==Proof== | ==Proof== | ||
| − | {{Requires proof|grade=C|msg=Proof isn't that important as it is easy and routine.}} | + | * Let {{M|\mathcal{U} }} be an [[open cover]] of {{M|f(A)}} and then showing it has a finite subcover |
| + | ** The proof depends on {{M|\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} }} being an [[open cover]] of {{M|A}} by sets [[open set|open]] in {{Top.|X|J}} | ||
| + | **# The open part comes from the {{M|U}} be [[open sets]], then by the definition of [[continuity]] {{M|f^{-1}(U)}} is open in {{M|X}} for each {{M|U\in\mathcal{U} }} | ||
| + | **# We need to show {{MM|A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)}}, by the [[implies-subset relation]] and then the definition of [[union]] we see: | ||
| + | **#* {{MM|\left(A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right)\iff\forall x\in A\left[x\in\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right]}} {{MM|\iff\forall x\in A\exists U\in\mathcal{U}[x\in f^{-1}(U)]}} | ||
| + | **#* In this case {{M|\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} }} is an [[open cover]] of {{M|A}} | ||
| + | ** Then we see, by [[compactness]] of {{M|A}}: | ||
| + | *** there is some {{M|n\in\mathbb{N} }} such that {{M|\{f^{-1}(U_1),\ldots,f^{-1}(U_n)\} }} | ||
| + | {{Requires proof|grade=C|msg=Proof isn't that important as it is easy and routine. | ||
| + | * Done a chunk, but not finished. [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 17:20, 18 December 2016 (UTC)}} | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Topology navbox|plain}} | {{Topology navbox|plain}} | ||
{{Theorem Of|Topology}} | {{Theorem Of|Topology}} | ||
Latest revision as of 17:20, 18 December 2016
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Flesh out with references, proof is small but easy and can wait
Contents
[hide]Statement
Let (X,J) and (Y,K) be topological spaces, let A∈P(X) be an arbitrary subset of X and let f:X→Y be a continuous map. Then:
- if A is compact then f(A) is compact.
Grade: A
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I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?
Proof
- Let U be an open cover of f(A) and then showing it has a finite subcover
- The proof depends on {f−1(U) | U∈U} being an open cover of A by sets open in (X,J)
- The open part comes from the U be open sets, then by the definition of continuity f−1(U) is open in X for each U∈U
- We need to show A⊆⋃U∈Uf−1(U), by the implies-subset relation and then the definition of union we see:
- (A⊆⋃U∈Uf−1(U))⟺∀x∈A[x∈⋃U∈Uf−1(U)]⟺∀x∈A∃U∈U[x∈f−1(U)]
- In this case {f−1(U) | U∈U} is an open cover of A
- (A⊆⋃U∈Uf−1(U))⟺∀x∈A[x∈⋃U∈Uf−1(U)]
- Then we see, by compactness of A:
- there is some n∈N such that {f−1(U1),…,f−1(Un)}
- The proof depends on {f−1(U) | U∈U} being an open cover of A by sets open in (X,J)
Grade: C
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References
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