Subset of Particular Point Space is either Open or Closed

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Theorem

Let $T = \left({S, \tau_p}\right)$ be a particular point space.

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


Then $H$ is either open or closed in $T$.


The only sets which are both closed and open in $T$ are $S$ and $\varnothing$.


Proof

Let $H \subseteq S$.

There are two cases to consider:

$p \in H$
$p \notin H$


If $p \in H$ then by definition of a particular point topology, $H$ is open.


If $p \notin H$, then $p \in \complement_S \left({H}\right)$, where $\complement_S \left({H}\right)$ is the relative complement of $H$ in $S$.

So $\complement_S \left({H}\right)$ is open by definition of a particular point topology.

From the definition of a closed set, $\complement_S \left({\complement_S \left({H}\right)}\right)$ is closed in $T$.

From Relative Complement of Relative Complement we have that $\complement_S \left({\complement_S \left({H}\right)}\right) = H$ and so $H$ is closed in $T$.


Now suppose $H \subseteq T$ is both closed and open in $T$.

From Open and Closed Sets in Topological Space, if $H = \varnothing$ or $H = S$ then H is both closed and open in $T$.

So, suppose $H \ne \varnothing$ or $H \ne S$.

As $H \subseteq T$ is both closed and open in $T$, $\complement_S \left({H}\right)$ is also both closed and open in $T$.

From Boundary is Intersection of Closure with Closure of Complement we have:

$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^-$

where $H^-$ is the closure of $H$.

By Closure of Open Set of Particular Point Space we have that $H^- = S$, and of course $\left({\complement_S \left({H}\right)}\right)^- = S$.

So:

$\partial H = H^- \cap \left({\complement_S \left({H}\right)}\right)^- = S$

However, from Set Clopen iff Boundary is Empty we have that $\partial H = \varnothing$.

Hence if $H$ is both closed and open in $T$, then $H = \varnothing$ or $H = S$.

$\blacksquare$