Topology Defined by Closed Sets

Theorem
Let $X$ be any set and let $\tau$ be a collection of subsets of $X$.

Then $\tau$ is a topology on $X$ iff:


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

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

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

Then by De Morgan's Laws, we have:
 * $\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \in \mathbb V} \left({X \setminus V}\right)$

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

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

Thus the properties as listed above hold.

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

all hold.

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

So $\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \in \mathbb V} \left({X \setminus V}\right)$ is open.

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

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

And of course by Open and Closed Sets in Topological Space, $\varnothing$ and $X$ are both open in $X$.

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