Topologies are not necessarily Comparable by Coarseness

Theorem
Let $S$ be a set with at least $2$ elements.

Let $\mathbb T$ be the set of all topologies on $S$.

For two topologies $\tau_a, \tau_b \in \mathbb T$, let $\tau_a \le \tau_b$ denote that $\tau_a$ is coarser than $\tau_b$.

Then there exist $\tau_1, \tau_2 \in \mathbb T$ such that neither:
 * $\tau_1 \le \tau_2$

nor:
 * $\tau_2 \le \tau_1$

That is, there are always topologies on $S$ which are non-comparable.

Proof
Let $a, b \in S$.

Let:
 * $\tau_a$ be the particular point topology with respect to $a$ on $S$
 * $\tau_b$ be the particular point topology with respect to $b$ on $S$

Then:
 * $\left\{{a}\right\} \in \tau_a$ but $\left\{{a}\right\} \notin \tau_b$
 * $\left\{{b}\right\} \in \tau_b$ but $\left\{{b}\right\} \notin \tau_a$

So neither $\tau_a \le \tau_b$ nor $\tau_b \le \tau_a$.