Open Set in Partition Topology is also Closed

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space.

Then $T$ is a partition space if and only if:

$\forall U \subseteq S: U \in \tau \iff \complement_S \left({U}\right) \in \tau$

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


Proof

Let $T = \left({S, \tau}\right)$ be a topological space.


Necessary Condition

Let $T$ be a partition space.

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

Let $U \in \tau$.

Then $U$ is the union of elements of $\mathcal P$.

Then by definition $\complement_S \left({U}\right)$ is closed in $T$.

Then $\complement_S \left({U}\right)$ is the union of all the other elements of $\mathcal P$.

That is, $\complement_S \left({U}\right)$ is also open in $T$.

By definition, then, $\complement_S \left({\complement_S \left({U}\right)}\right) = U$ is closed in $T$.

$\Box$


Sufficient Condition

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

Aiming for a contradiction, suppose $T$ is not a partition space.

Thus, by definition, there exists an open set of $T$ which is not the union of disjoint sets.



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