Either-Or Topology is Scattered

Source Work

 * Part $\text{II}$: Counterexamples
 * Section $17.5:$ Either-Or Topology
 * Section $17.5:$ Either-Or Topology

This mistake can be seen in the second edition (1978) as republished by Dover in 1995: ISBN 0-486-68735-X

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

Then $T$ is a scattered space.


 * "However $X$ is scattered, since there are no nonempty dense-in-itself subsets, for $0$ is the only possible limit point of any subset."

Supposed Proof
Let $x \in T, x \ne 0$.

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

So $\left\{{x}\right\}$ is a neighborhood of $x$ containing only $x$, and so $x$ is isolated.

So a subset of $H \subseteq X$ contains isolated points and so by definition is not dense-in-itself.

So by definition $T$ is a scattered space.

Analysis
The above proof appears to be flawed, as the set $\left\{{0}\right\}$ is contained in the open set $\left({-1 .. 1}\right)$ and it follows that $0$ is not isolated.

Hence $\left\{{0}\right\}$ is dense-in-itself, and so $T$ is, according to this analysis, not scattered after all.