Homeomorphic Non-Comparable Particular Point Topologies

Theorem
Let $S$ be a set with at least two elements.

Let $p, q \in S: p \ne q$.

Let $\tau_p$ and $\tau_q$ be the particular point topologies on $S$ by $p$ and $q$ respectively.

Then the topological spaces $T_p = \left({S, \tau_p}\right)$ and $T_q = \left({S, \tau_q}\right)$ are homeomorphic.

However, $\tau_p$ and $\tau_q$ are not comparable.

Proof
We can set up the mapping $\phi: S \to S$:
 * $\forall x \in S: \phi \left({x}\right) = \begin{cases}

q & : x = p \\ p & : x = q \\ x & : \text {otherwise} \end{cases}$

It is straightforward to show that $\phi$ is a homeomorphism.

However, we have, for example, that $\left\{{q}\right\} \notin \tau_p$ and $\left\{{p}\right\} \notin \tau_q$.

So neither $\tau_p$ nor $\tau_q$ are a subset of the other.

Hence by definition $\tau_p$ and $\tau_q$ are not comparable.