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) Any intersection of arbitrarily many closed sets of $$X$$ under $$\vartheta$$ is a closed set of $$X$$ under $$\vartheta$$;
 * 2) The union of any finite number of closed sets of $$X$$ under $$\vartheta$$ is a closed set of $$X$$ under $$\vartheta$$;
 * 3) $$X$$ 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:
 * $$X \setminus \bigcap \mathbb V = \bigcup_{V \in \mathbb V} \left({X \setminus V}\right)$$

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

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

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

By a similar argument, if $$\bigcap_{i=1}^n V_i$$ is the union of some finite number of closed sets of $$X$$, it follows that $$\bigcap_{i=1}^n V_i$$ 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: all hold.
 * 1) Any intersection of arbitrarily many closed sets of $$X$$ under $$\vartheta$$ is a closed set of $$X$$ under $$\vartheta$$;
 * 2) The union of any finite number of closed sets of $$X$$ under $$\vartheta$$ is a closed set of $$X$$ under $$\vartheta$$;
 * 3) $$X$$ and $$\varnothing$$ are both closed sets of $$X$$ under $$\vartheta$$.

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

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