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.