The image of a compact set is compact

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Stub grade: A
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Flesh out with references, proof is small but easy and can wait

Statement

Let (X,J) and (Y,K) be topological spaces, let AP(X) be an arbitrary subset of X and let f:XY 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 {f1(U) | UU} being an open cover of A by sets open in (X,J)
      1. The open part comes from the U be open sets, then by the definition of continuity f1(U) is open in X for each UU
      2. We need to show AUUf1(U)
        , by the implies-subset relation and then the definition of union we see:
        • (AUUf1(U))xA[xUUf1(U)]
          xAUU[xf1(U)]
        • In this case {f1(U) | UU} is an open cover of A
    • Then we see, by compactness of A:
      • there is some nN such that {f1(U1),,f1(Un)}
Grade: C
This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.
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Proof isn't that important as it is easy and routine.
  • Done a chunk, but not finished. Alec (talk) 17:20, 18 December 2016 (UTC)

References