The image of a compact set is compact
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Flesh out with references, proof is small but easy and can wait
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[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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