Limit Points of Either-Or Topology

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $H \subseteq S$ be any subset of $S$.


Then no element of $S$ can be a limit point of $H$ except $0$.


Proof

Let $x \in S$ such that $x \ne 0$.

Then, as $0 \notin \left\{{x}\right\}$, we have by definition of the either-or topology that $x$ is open in $T$.

So whatever $H$ is, $\left\{{x}\right\}$ never contains any points of $H$ which are different from $x$.

So $x$ can not be a limit point of $H$.


However, every open set of $T$ which contains $0$ also contains the interval $\left({-1 \,.\,.\, 1}\right)$ and so can contain points of $H$.

It follows that the only non-empty sets of $T$ for which $0$ is not a limit point are $\left\{{-1}\right\}$, $\left\{{0}\right\}$ and $\left\{{1}\right\}$ and their unions.

$\blacksquare$


Sources