# Either-Or Topology is Scattered

## Theorem

Let $T = \struct {S, \tau}$ 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, $\set x$ is open in $T$.

So $\set x$ 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: $H = \set 0$.

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.

$\blacksquare$

## 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: $5$