Compact Sets, Hausdorff Spaces

Maps from Compact to Hausdorff

If S is compact in a hausdorff space, it is closed. Consider any point x not in S. For each y in S, embed y in an open set disjoint from an open set containing x. Do this for each y in S, and we have an open cover. Take a finite subcover, intersect the corresponding open sets about x, and find an open set that contains x, and misses S. This holds for all x outside of S, hence the complement of S is open, and S is closed.

An important corollary is that a continuous 1-to-1 map of a compact space onto a hausdorff space is bicontinuous. Any closed set in the domain is also compact, its continuous image is compact, and its image is closed. Taking complements, the image of open sets is open. Thus the function is bicontinuous. Inother words, a homeomorphism. The two spaces are topologically equivalent.

The word onto is important here. Embed the closed interval [0,1] into the real line, and the result is not bicontinuous. The interval [0,½) is open in the domain, but its image is not open in the range.

Let f map X onto Y and let g map Y into Z. Let X be compact and let Y be Hausdorff. Let the functions f and g(f) be continuous. Start with a closed set in Z and take its preimage under g. If this preimage is not closed in Y, the prepreimage cannot be closed in X, since f maps closed sets to closed sets. Yet f(g) is continuous, hence both preimages are closed, and g is continuous.

Let S be hausdorff and let U be a compact subspace of S. If U is also dense in S then U = S. Remember that a compact set in a hausdorff space is closed, so U is closed. If U misses the point x then U must miss some open set containing x, which contradicts the fact that U is dense. Thus U contains every x, and U = S.

Compact + Hausdorff Walks a Fine Line

Making a topology weaker keeps a space compact, for any open cover is an open cover in the original compact space. In contrast, making a topology stronger, i.e. adding more open sets, keeps a space hausdorff.

Let S be a space that is compact and hausdorff. watch what happens to S with a stronger or weaker topology. Let f map S onto itself, but let the domain have a stronger topology. Open sets still have open sets as preimage, so f is continuous. Yet some open sets in the domain are new, and don't map to open sets in S, hence f is not bicontinuous. We already showed that a map from a compact space onto a hausdorff space is a homeomorphism, and that isn't the case here; hence the domain of f is not compact. Similarly, when g maps S onto itself with a weaker topology, g is not bicontinuous, and the range is not hausdorff.

A compact hausdorff space walks a fine line. Add more open sets and lose compactness. Take away open sets and the space is no longer hausdorff.

Compact + Hausdorff is Normal

Assume a space is both compact and hausdorff. Let S be a closed set in this space, and let x be any point not in S. Separate every point in S from x, using disjoint open sets. A finite cover will do, so we can separate S from x, and the space is regular.

Let T be another closed set, disjoint from S. For every x in T, separate x and S in disjoint open sets. Again, a finite cover will do. A finite intersection of open sets contains S, while a finite union of open sets contains T, and the space is normal.

Compact + Hausdorff Embeds in a Hypercube

If S is compact and hausdorff, a homeomorphism embeds S in a hypercube. For all practical purposes, S is a subspace of the cube.

Consider all possible continuous functions from S into [0,1]. These form a composite function f that maps S into the product space P, a hypercube with the weak product topology.

Select any pair of points x and y in S. Because S is normal, Urysohn's lemma provides a function fj that maps x to 0 and y to 1. Since x and y attain different values on at least one coordinate, f(x) ≠ f(y), and f is 1-1. In other words, f embeds S into P, and the map is continuous. Since the domain is compact and the range hausdorff, f is a homeomorphism.