# Compactness and Hausdorffness with different topology

Here is the question (Munkres pg. 170):

Show that if $X$ is compact Hausdorff under both $\mathcal{T}$ and $\mathcal{T}'$, then either $\mathcal{T}$ and $\mathcal{T}'$ are equal or they are not comparable.

The solution I found is that:

Suppose one is finer than the other. Then the identity mapping from the finer one to the coarser one is a continuous and bijective function that maps a compact space to a Hausdorff space. Therefore, it is a homeomorphism and the topologies are the same.

I don't know where is the "not comparable" part. I tried to create examples in lower limit topology and and K-topology, but it does not make things clearer. Can someone help with the incomparablility?

Incomparable here simply means that neither is a subset of the other: neither is finer or coarser than the other. Specifically, the proof shows that if $\tau$ and $\tau'$ are different compact Hausdorff topologies on $X$, then $\tau\nsubseteq\tau'$ and $\tau'\nsubseteq\tau$, and this is precisely what is meant by saying $\tau$ and $\tau'$ incomparable.
For an example of incomparable compact Hausdorff topologies on the same set, let $X=[0,1]$, and let $\tau$ be the usual Euclidean topology on $X$. Let $C$ be the middle-thirds Cantor set; $|C|=\mathscr{c}=|[0,1]|$, so there is a bijection $h:C\to[0,1]$. Let $\tau'=\{h[U]:U\text{ is open in }C\}$; then $\tau'$ is a topology on $X$, and $h$ is a homeomorphism of $C$ onto $\langle X,\tau'\rangle$, which is therefore compact and Hausdorff. Finally, $C$ is not connected, while $[0,1]$ is, so $\tau$ and $\tau'$ are clearly not the same topology.
• @lomber You can give separation of C as C=$C\cap (-1,0.5) \cup C\cap (0.5,2)$ – MathLover Mar 26 at 11:31