Union of Topologies is not necessarily Topology
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Theorem
Let $\tau_1$ and $\tau_2$ be topologies on a set $S$.
Then $\tau_1 \cup \tau_2$ is not necessarily also a topology on $S$.
Proof
Let $S := \set {0, 1, 2}$ be a set.
Let:
- $\tau_1 := \set {\O, \set 0, \set 1, \set {0, 1}, S}$
- $\tau_2 := \set {\O, \set 0, \set 2, \set {0, 2}, S}$
be topologies on $S$.
Then:
- $\tau := \tau_1 \cup \tau_2 = \set {\O, \set 0, \set 1, \set 2, \set {0, 1} \set {0, 2}, S}$
For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$.
But:
- $\set 1 \cup \set 2 = \set {1, 2} \not \in \tau$
Therefore $\tau$ is not a topology on $S$.
Hence the result.
$\blacksquare$
Also see
- Union of Topologies on Singleton or Doubleton is Topology where it is shown that this result does not hold if $\size S \le 2$.
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: Exercise $3.9: 4$