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?
 A: Two topologies are "comparable" if one is finer than the other, i.e. every open set in one is open in the other.  As the solution shows, if this is the case the two topologies must be the same.
A: 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.
