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

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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 $E$