Either-Or Topology is Scattered

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Theorem

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

Then $T$ is a scattered space.


Proof

By definition, $T$ is scattered if and only if every non-empty subset $H$ of $S$ contains at least one point which is isolated in $H$.


Let $H$ be a non-empty subset of $S$.

Let $x \in H$.

Let $x \ne 0$.

By definition of either-or space, $\left\{{x}\right\}$ is open in $T$.

So $\left\{{x}\right\}$ is an open set of $x$ containing only $x$.

Thus by definition $x$ is isolated.

Thus if $H$ is a non-empty subset of $S$ which contains any element other than $0$, $H$ contains at least one isolated point.


The other case is that $H$ contains no $x$ such that $x \ne 0$.

That is: $H = \left\{{0}\right\}$.

But from Singleton Point is Isolated, $0$ is an isolated point in $H$.

Thus in all cases $H$ contains at least one point which is isolated in $H$.

So by definition $T$ is a scattered space.

$\blacksquare$


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