Open Set in Partition Topology is also Closed

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

Then $T$ is a partition space :
 * $\forall U \subseteq S: U \in \tau \iff \relcomp S U \in \tau$

That is, a topological space is a partition space all open sets are closed, and all closed sets are also open.

Proof
Let $T = \struct {S, \tau}$ be a topological space.

Necessary Condition
Let $T$ be a partition space.

Then there exists a partition $\PP$ which forms the basis for $T$.

Let $U \in \tau$.

Then $U$ is the union of elements of $\PP$.

Then by definition $\relcomp S U$ is closed in $T$.

Then $\relcomp S U$ is the union of all the other elements of $\PP$.

That is, $\relcomp S U$ is also open in $T$.

By definition, then, $\relcomp S {\relcomp S U} = U$ is closed in $T$.

Sufficient Condition
Now suppose that every open set in $T$ is closed, and every closed set in $T$ is open.

For each $x \in S$, define:
 * $\ds U_x := \bigcap_{x \mathop \in U \mathop \in \tau} U$

We claim that $\Bbb S = \set {U_x: x \in S}$ is a partition of $S$.

To show that this is true, we have to check the following:
 * $(1): \quad \Bbb S$ is pairwise disjoint: $\forall S_1, S_2 \in \Bbb S: S_1 \cap S_2 = \O$ when $S_1 \ne S_2$
 * $(2): \quad$ The union of $\Bbb S$ forms the whole set $S$: $\ds \bigcup \Bbb S = S$
 * $(3): \quad$ None of the elements of $\Bbb S$ is empty: $\forall A \in \Bbb S: A \ne \O$.

Clearly $x \in U_x \ne \O$ and $\ds S \subseteq \bigcup \Bbb S \subseteq S$, so we only need to show that $\Bbb S$ is pairwise disjoint.

Since each $U \in \tau$ is closed, each $U_x$ is an intersection of closed sets.

By Intersection of Closed Sets is Closed in Topological Space, each $U_x$ is also closed.

Since every closed set in $T$ is open, we have $U_x \in \tau$.

Pick any $a, b \in S$.

Suppose $U_a \ne U_b$.

Note that $U_a \cap U_b$ is also closed, so $U_a \cap U_b \in \tau$.

If $a \notin U_a \cap U_b$, we have $a \in U_a \setminus \paren {U_a \cap U_b}$.

By Open Set minus Closed Set is Open, $U_a \setminus \paren {U_a \cap U_b} \in \tau$.

By Intersection is Subset and Set Difference is Subset:
 * $\ds U_a = \bigcap_{a \mathop \in U \mathop \in \tau} U \subseteq U_a \setminus \paren {U_a \cap U_b} \subseteq U_a$

so we have $U_a = U_a \setminus \paren {U_a \cap U_b}$ by definition of set equality.

By Set Difference Equals First Set iff Empty Intersection, we have $U_a \cap U_b = \O$.

We get a similar result if $b \notin U_a \cap U_b$.

Now suppose $a, b \in U_a \cap U_b$.

By Intersection is Subset:
 * $\ds U_a = \bigcap_{a \mathop \in U \mathop \in \tau} U \subseteq U_a \cap U_b \subseteq U_a$

so we have $U_a \cap U_b = U_a$ by definition of set equality.

Similarly we have $U_a \cap U_b = U_b$.

Hence $U_a = U_b$, which is a contradiction.

Therefore:
 * $U_a \ne U_b \implies U_a \cap U_b = \O$

and thus $\Bbb S$ is a partition of $S$.

Finally we need to show that each $V \in \tau$ is a union of sets in $\Bbb S$.

In particular, we will show:
 * $\ds V = \bigcup_{x \mathop \in V} U_x$

Clearly:
 * $\ds V \subseteq \bigcup_{x \mathop \in V} U_x$

For the other inclusion, by Intersection is Subset we have:
 * $\ds \forall x \in V: U_x = \bigcap_{x \mathop \in U \mathop \in \tau} U \subseteq V$

By Union of Subsets in Subset:
 * $\ds \bigcup_{x \mathop \in V} U_x \subseteq V$

Hence by definition of set equality:
 * $\ds V = \bigcup_{x \mathop \in V} U_x$

And thus $T$ is a partition space.