Topology Defined by Closed Sets

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

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


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

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 (Set Theory), we have:
 * $\displaystyle X \setminus \bigcap \mathbb V = \bigcup_{V \in \mathbb V} \left({X \setminus V}\right)$

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

As $X \setminus V$ is open it follows that $\displaystyle \bigcup_{V \in \mathbb V} \left({X \setminus V}\right)$ is open by the definition of topology.

Thus $\displaystyle X \setminus \bigcap \mathbb V$ is open and by definition $\displaystyle \bigcap \mathbb V$ is closed.

By a similar argument, if $\displaystyle \bigcap_{i=1}^n V_i$ is the union of some finite number of closed sets of $X$, it follows that $\displaystyle \bigcap_{i=1}^n V_i$ is closed.

Finally note that by Open and Closed Sets in a 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 $\vartheta$ is a closed set of $X$ under $\vartheta$
 * $(2): \quad$ The union of any finite number of closed sets of $X$ under $\vartheta$ is a closed set of $X$ under $\vartheta$
 * $(3): \quad X$ and $\varnothing$ are both closed sets of $X$ under $\vartheta$.

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 $\vartheta$ is an open set of $X$ under $\vartheta$.

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

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

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