Home > Functional Analysis, Topology > Compact & Hausdorff spaces

Compact & Hausdorff spaces

The following problem wants to prove that compact and Hausdorff spaces have an interesting property. If (X,\mathcal{T}) is Hausdorff and compact, and we consider another Hausdorff topology \mathcal{T}^\prime coarser than \mathcal{T} and Hausdorff, then \mathcal{T}=\mathcal{T}^\prime.

1) Prove that a compact set in a Hausdorff space is closed.

2) Every continuous function from a compact space onto a Hausdorff space is open.

3) If you have a bijective continuous function from a compact space to a Hausdorff space, then the two spaces are homeomorphic.

A short application, which was my motivation for this post is the following problem:

Let E be a Banach space and K\subset E be a compact subset in the strong topology. Let (x_n) be a sequence in K such that x_n\to x in the weak topology. Prove that x_n \to x strongly.

Proof: 1) Suppose S\subset X is a compact set. We can assume that S \neq X. Pick x \notin S. For every y there exists an open set O_y which does not intersect an open set O \ni x. Doing this for every y \neq x, y \in S with O we get an open cover of S, which has a finite subcover O_{y_1},...,O_{y_k}. Then for the open set not containing x, \bigcup_j O_{y_j} we can find an open set containing x which has nothing in common with this union. The union contains S, and therefore x is contained in an open set outside S. Doing this for every x \notin S we get the desired result.

2) Any closed set is compact, which is mapped by f into a compact set, and by 1) this is closed. Taking complements we see that the image of any open set is open. (at this point we have used that f is onto!)

3) f is invertible and use 2).

To prove the first statement that any coarser Hausdorff topology is the same as the initial topology, we proceed as follows:

Take I: (X,\mathcal{T}) \to (X,\mathcal{T}^\prime), where (X,\mathcal{T}) is compact and Hausdorff and (X,\mathcal{T}^\prime) is also Hausdorff, and I is the identity mapping. This is a continuous mapping since \mathcal{T}^\prime \subset \mathcal{T}. It is also bijective, and therefore a homeomorphism. Therefore the two spaces have the same topological structure and \mathcal{T}=\mathcal{T}^\prime.

The intuitive idea is that adding open sets preserves Hausdorff; removing open sets preserves compactness and if you want to keep both you must leave them as they are. 🙂

The short problem at the end is proven already in the above steps. The weak topology is coarser than the strong topology and Hausdorff, and K is a compact space. Therefore the topologies are the same.

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