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
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $17$. Either-Or Topology