# A topology is closed under finite union and arbitrary intersection of closed sets.

Show that a topology is closed under finite union and arbitrary intersection of closed sets.

There are some details that are bothering me in this question. Let $$(E,\tau)$$ be a topological space. We say that $$A\subseteq E$$ is closed under $$\tau$$ iff $$A^c \in \tau$$. I know there are other definitions than that where each element of $$\tau$$ is open, but this latter is the only which was presented to me so far. So I am supposed to work with it.

In this context, to show the proposition, that 'closed sets' in the question are actually "closed sets under $$\tau$$" right? For a given choice of $$\tau$$, I don't know if it is guaranteed that the complement of any closed set in $$E$$ is in $$\tau$$.

Furthermore, given $$\tau$$, is it right to say that $$\tau$$ is "closed under ... of closed sets"? As far as I understand, I was asked to prove that if $$\tau$$ is a topology on $$E$$ then the finite union and arbitrary intersection of closed sets of $$E$$ under $$\tau$$ is closed sets of $$E$$ under $$\tau$$. It does not mean that $$\tau$$ is closed under these operations of closed sets. The notion of a set being "closed under intersections or unions" is that for every element in these set, the intersection or unions between them are in the set. But all elements of $$\tau$$ is open.

Can you clarify these points to me?

Thanks

Yes, it is guaranteed that the complement of any closed set in $$(E,\tau)$$ is in $$\tau$$, since, by definition, asserting that $$A$$ is closed means that $$A^\complement\in\tau$$.

On the other hand, it makes no sense to assert that $$\tau$$ is “closed under … of closed sets”. But that is not the question. There are in fact two questions:

• the set of closed subsets of $$(E,\tau)$$ is closed under finite unions;
• the set of closed subsets of $$(E,\tau)$$ is closed under arbitrary intersections.

A better formulation would have been: the collection of closed sets of $$(X,\tau)$$ are closed under finite unions and also closed under arbitrary intersections. Often instead of the complete (and somewhat formal) $$(X,\tau)$$, the shorthand $$X$$ (or less commonly $$\tau$$) is used, which can be confusing, as we see.

The fact itself is immediate by de Morgan's laws for unions/intersections and complements, e.g.:

If $$F_1, F_2, \ldots F_n$$ are all closed then $$F_1^\complement, F_2^\complement, \ldots, F_n^\complement$$ are all open and so $$\bigcap_{i=1}^n F_i^\complement$$ is open by an axiom of topologies and so its complement $$\left(\bigcap_{i=1}^n F_i^\complement\right)^\complement$$ is closed and this set equals $$\bigcup_{i=1}^n F_i$$ by de Morgan. The intersections are shown similarly.

You are correct in your statements about closed sets in $$E$$ with respect to the topology $$t$$

Once a topology is defined on $$E$$ we have the open sets in $$E$$ and their complements are closed in $$E$$