Continuous image of a compact space

The proof is direct: if a family of open sets of \(Y\) covers \(f(X)\), their preimages are open sets of \(X\) (by continuity) covering \(X\); compactness leaves finitely many of them, and the images of those finitely many preimages still cover \(f(X)\).

The compactness hypothesis falls on the domain. The continuous function \(f(x)=1/x\) sends the noncompact \((0,1]\) to \([1,\infty)\), which is also noncompact; and a continuous function from a noncompact domain can have compact image, as in \(\sin:\mathbb{R}\to[-1,1]\). The converse is therefore false.

In particular, a continuous \(f:X\to\mathbb{R}\) with \(X\) compact has compact image in \(\mathbb{R}\), that is, closed and bounded: it is bounded and attains its maximum and minimum. This is the extreme value theorem, which justifies looking for extrema of \(f\) on \([a,b]\) among the endpoints and the critical points.

Another consequence: compactness is a topological property, since if \(f\) is a homeomorphism then \(f(X)\) and \(X\) are compact simultaneously. Thus \((0,1)\) is not homeomorphic to \([0,1]\).