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 $S$

From Homeomorphic Non-Comparable Particular Point Topologies:

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

Hence the result.

$\blacksquare$

Also see

• Coarseness Relation on Topologies is Partial Ordering

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.1$: Topological Spaces
• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 3$
• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction