Either-Or Topology is Topology

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $T = \left({S, \tau}\right)$ be the either-or space.


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


Proof

From the definition:

$H \in \tau \iff \left({\left\{{0}\right\} \nsubseteq H \lor \left({-1 \,.\,.\, 1}\right) \subseteq H}\right)$

for any $H \subseteq S$.

First note that:

$\left\{{0}\right\} \nsubseteq \varnothing$ and so $\varnothing \in \tau$
$\left({-1 \,.\,.\, 1}\right) \subseteq S$ and so $S \in \tau$


Now suppose $H_1, H_2 \in \tau$.

If either $\left\{{0}\right\} \nsubseteq H_1$ or $\left\{{0}\right\} \nsubseteq H_2$ then from De Morgan's Laws: Difference with Intersection (indirectly):

$\left\{{0}\right\} \nsubseteq H_1 \cap H_2$

Otherwise:

$\left({-1 \,.\,.\, 1}\right) \subseteq H_1$ and $\left({-1 \,.\,.\, 1}\right) \subseteq H_2$

and so from Intersection is Largest Subset:

$\left({-1 \,.\,.\,1}\right) \subseteq H_1 \cap H_2$

So:

$H_1, H_2 \in \tau \implies H_1 \cap H_2 \in \tau$


Now let $\mathcal H \subseteq \tau$ be a set of elements of $\tau$.

Then either:

$\forall H \in \mathcal H: \left\{{0}\right\} \nsubseteq H$ in which case $\left\{{0}\right\} \nsubseteq \bigcup \mathcal H$

or:

$\exists H \in \mathcal H: \left({-1 \,.\,.\, 1}\right) \subseteq H$ in which case $\left({-1 \,.\,.\, 1}\right) \subseteq \bigcup \mathcal H$

In both cases $\displaystyle \bigcup \mathcal H \in \tau$.


Hence the result, from the definition of topology.

$\blacksquare$


Note that $\left\{{0}\right\} \nsubseteq H \implies \left({-1 \,.\,.\, 1}\right) \nsubseteq H$ so there does not exist the confusion of what happens if the conditions are contradictory.


Sources