Compactness is Preserved Under Continuous Functions

One key topology homeomorphism invariant is compactness, a fact fundamental to the extreme value theorem.

Theorem

Let f:XY be a continuous function between topological spaces X and Y, and let X be a compact subset of X. Then, f(X) is compact as a subset of Y

Proof

The main strategy here is to take an arbitrary open cover of f(X), use continuity with images to induce an open cover of X, which has a finite subcover, which can then be used to find a corresponding finite subcover of f(X) in the codomain.

Consider an arbitrary open cover of f(X), that is

iIVif(X).

Then, since pre-images preserve unions,

f1(iIVi)=iIf1(Vi).

First we show that this set is an open cover of X.

Each set is open from continuity of f and openness of Vi for each i.

Then because pre-images preserve inclusions and any set is a subset of the pre-image of its image under some function, we have that

iIVif(X)f1(iIVi)f1(f(X))X

and hence the desired set covers X, this cover has a finite subcover, namely a finite subset JI such that

jJf1(Vj)

is an open cover of X.

We will now prove that the corresponding indices give a subcover of f(X), that is that

jJVj

is a finite cover of f(X) which is a subcover of our original cover.

Clearly this cover is open, and finite, we simply must prove that it does indeed cover f(X).

Now, for each yf(X), there exists and xX such that f(x)=y. Then, since jJf1(Vj) covers X, we know that xVj for some jJ and hence f(x)=yf(Vj) for the same j. This means for any yf(X), our finite subcover contains y, and thus it covers f(X).