Topologies are not necessarily Comparable by Coarseness
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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
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