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
Best Answer
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: