Closed Sets of Either-Or Topology

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Theorem

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


Then the closed sets of $T$ are:

  • $\varnothing$
  • $S$
  • $\left\{{-1}\right\}$
  • $\left\{{1}\right\}$
  • $\left\{{-1, 1}\right\}$
  • Any subset of $\left[{-1 \,.\,.\, 1}\right]$ containing $\left\{{0}\right\}$ as a subset.


Proof

From Open and Closed Sets in Topological Space we have that $\varnothing$ and $S$ are closed sets trivially.


From the definition of closed set, we have:

$U$ is open in $T$ if and only if $S \setminus U$ is closed in $T$
$U$ is closed in $T$ if and only if $S \setminus U$ is open in $T$

where $S \setminus U$ denotes the relative complement of $U$ in $S$.

Now we have that:

$\left({-1 \,.\,.\, 1}\right)$ is open in $T$ so $S \setminus \left({-1 \,.\,.\, 1}\right) = \left\{{-1, 1}\right\}$ is closed in $T$.
Both $\left[{-1 \,.\,.\, 1}\right) \supseteq \left({-1 \,.\,.\, 1}\right)$ and $\left({-1 \,.\,.\, 1}\right] \supseteq \left({-1 \,.\,.\, 1}\right)$ therefore are open in $T$, so $\left\{{-1}\right\}$ and $\left\{{1}\right\}$ are closed in $T$.

Apart from $S$ itself, there are no other subsets of $S = \left[{-1 \,.\,.\, 1}\right]$ which have $\left({-1 \,.\,.\, 1}\right)$ as a subset.


Finally, let $U \in \tau: \left\{{0}\right\} \nsubseteq U$.

Then by the definition of relative complement $\left\{{0}\right\} \subseteq S \setminus U$.

Now suppose $\left\{{0}\right\} \subseteq V$.

Then again by the definition of relative complement $\left\{{0}\right\} \nsubseteq S \setminus V$.

So $S \setminus V$ is open in $T$ and so $V$ is closed in $T$.

$\blacksquare$


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