Topology Defined by Closed Sets

Theorem
Let $S$ be a set.

Let $\tau$ be a set of subsets of $S$.

Then $\tau$ is a topology on $S$ :


 * $(1): \quad$ Any intersection of arbitrarily many closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
 * $(2): \quad$ The union of any finite number of closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
 * $(3): \quad S$ and $\O$ are both closed sets of $S$ under $\tau$

where a closed set $V$ of $S$ under $\tau$ is defined as a subset of $S$ such that $S \setminus V \in \tau$.

Proof
From the definition, if $V$ is a closed set of $S$, then $S \setminus V$ is an open set of $S$.

Let $\mathbb V$ be any arbitrary set of closed sets of $S$.

Then by De Morgan's Laws: Difference with Intersection, we have:
 * $\ds S \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {S \setminus V}$

First, let $\tau$ be a topology on $S$.

We have that:
 * Intersection of Closed Sets is Closed in Topological Space
 * Finite Union of Closed Sets is Closed in Topological Space
 * By Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed in $S$.

Thus, the properties as listed above hold.

Suppose the properties:
 * $(1): \quad$ Any intersection of arbitrarily many closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
 * $(2): \quad$ The union of any finite number of closed sets of $S$ under $\tau$ is a closed set of $S$ under $\tau$
 * $(3): \quad S$ and $\O$ are both closed sets of $S$ under $\tau$.

all hold.

That means $\ds \bigcap \mathbb V$ is closed.

So $\ds S \setminus \bigcap \mathbb V = \bigcup_{V \mathop \in \mathbb V} \paren {S \setminus V}$ is open.

Thus we have that the union of arbitrarily many open sets of $S$ under $\tau$ is an open set of $S$ under $\tau$.

Similarly, we deduce that the intersection of any finite number of open sets of $S$ under $\tau$ is an open set of $S$ under $\tau$.

By Open and Closed Sets in Topological Space, $\O$ and $S$ are both open in $S$.

So $\tau$ is a topology on $S$.