# The image of a compact set is compact

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:
Flesh out with references, proof is small but easy and can wait

## Statement

Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces, let [ilmath]A\in\mathcal{P}(X)[/ilmath] be an arbitrary subset of [ilmath]X[/ilmath] and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map. Then:

• if [ilmath]A[/ilmath] is compact then [ilmath]f(A)[/ilmath] is compact.
This page requires references, it is on a to-do list for being expanded with them.
Please note that this does not mean the content is unreliable, it just means that the author of the page doesn't have a book to hand, or remember the book to find it, which would have been a suitable reference.
The message provided is:
I want Mendelson. Also should I make a note that considered as a subspace or compactness-as-a-subset are the same?

## Proof

• Let [ilmath]\mathcal{U} [/ilmath] be an open cover of [ilmath]f(A)[/ilmath] and then showing it has a finite subcover
• The proof depends on [ilmath]\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} [/ilmath] being an open cover of [ilmath]A[/ilmath] by sets open in [ilmath](X,\mathcal{ J })[/ilmath]
1. The open part comes from the [ilmath]U[/ilmath] be open sets, then by the definition of continuity [ilmath]f^{-1}(U)[/ilmath] is open in [ilmath]X[/ilmath] for each [ilmath]U\in\mathcal{U} [/ilmath]
2. We need to show $A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)$, by the implies-subset relation and then the definition of union we see:
• $\left(A\subseteq\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right)\iff\forall x\in A\left[x\in\bigcup_{U\in\mathcal{U} }f^{-1}(U)\right]$ $\iff\forall x\in A\exists U\in\mathcal{U}[x\in f^{-1}(U)]$
• In this case [ilmath]\{f^{-1}(U)\ \vert\ U\in\mathcal{U} \} [/ilmath] is an open cover of [ilmath]A[/ilmath]
• Then we see, by compactness of [ilmath]A[/ilmath]:
• there is some [ilmath]n\in\mathbb{N} [/ilmath] such that [ilmath]\{f^{-1}(U_1),\ldots,f^{-1}(U_n)\} [/ilmath]