Either-Or Topology is Topology

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Theorem

Let $T = \struct {S, \tau}$ be the either-or topology.


Then $\tau$ is a topology on $T$.


Proof

Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets

Let $\UU \subseteq \tau$.

Then either:

$\forall U \in \UU: \set 0 \nsubseteq U$ in which case $\set 0 \nsubseteq \bigcup \UU$

or:

$\exists U \in \UU: \openint {-1} 1 \subseteq U$ in which case $\openint {-1} 1 \subseteq \bigcup \UU$

In both cases:

$\ds \bigcup \UU \in \tau$


Note that:

$\set 0 \nsubseteq U \implies \openint {-1} 1 \nsubseteq U$

so there does not exist the confusion of what happens if the conditions are contradictory.

$\Box$


Open Set Axiom $\paren {\text O 2 }$: Pairwise Intersection of Open Sets

Let $U_1, U_2 \in \tau$.

By definition of either-or topology, either:

$\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$

or:

$\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$


Suppose:

$\openint {-1} 1 \subseteq U_1$ and $\openint {-1} 1 \subseteq U_2$

From Intersection is Largest Subset:

$\openint {-1} 1 \subseteq U_1 \cap U_2$

By definition of either-or topology:

$U_1 \cap U_2 \in \tau$


Suppose:

$\set 0 \nsubseteq U_1$ or $\set 0 \nsubseteq U_2$

Then:

\(\ds \set 0\) \(\nsubseteq\) \(\ds U_1\)
\(\, \ds \lor \, \) \(\ds \set 0\) \(\nsubseteq\) \(\ds U_2\)
\(\ds \leadsto \ \ \) \(\ds \set 0\) \(\subseteq\) \(\ds \relcomp S {U_1}\) Definition of Relative Complement
\(\, \ds \lor \, \) \(\ds \set 0\) \(\subseteq\) \(\ds \relcomp S {U_2}\)
\(\ds \leadsto \ \ \) \(\ds \set 0\) \(\subseteq\) \(\ds \relcomp S {U_1} \cup \relcomp S {U_2}\) Definition of Set Union
\(\ds \leadsto \ \ \) \(\ds \set 0\) \(\subseteq\) \(\ds \relcomp S {U_1 \cap U_2}\) De Morgan's Laws: Difference with Intersection
\(\ds \leadsto \ \ \) \(\ds \set 0\) \(\nsubseteq\) \(\ds U_1 \cap U_2\) Definition of Relative Complement

By definition of either-or topology:

$U_1 \cap U_2 \in \tau$


Thus by Proof by Cases:

$\forall U_1, U_2 \in \tau: U_1 \cap U_2 \in \tau$

$\Box$


Open Set Axiom $\paren {\text O 3 }$: Underlying Set is Element of Topology

From Open Real Interval is Subset of Closed Real Interval:

$\openint {-1} 1 \subseteq \closedint {-1} 1 = S$

By definition of either-or topology:

$S \in \tau$

$\Box$


Hence the result, from the definition of topology.

$\blacksquare$


Sources

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