I need to prove that the intersection of any number of closed sets is a closed set itself. There are some proofs that come up upon doing some research but I want to do it with a very specific definition. Here it is:
A set A is closed if every point that is arbitrarily close to A is a member of A.
(Equivalently , if y –> x with y being an element of A, then x is an element of A)
I've been trying to figure out how to use this definition to prove that the intersection of any number of sets is closed as well.
Best Answer
Assuming (As this is not stated) that $x$ is "infinitely close" to a set $A$ means that every open set $O$ that contains $x$ intersects $A$ (or with open balls of radius $>0$ in a metric setting) (which is usually called that $x$ is an adherent point of $A$), the proof could go:
Suppose $A_i, i \in I$ are all closed in this sense. Define $A = \cap_i A_i$. Let $x$ be "close to" $A$. We want to show that $x \in A$. So suppose for a contradiction that $x \notin A$. Then there is some $i$ such that $x \notin A_i$. As $A_i$ is closed, this would mean that $x$ cannot be "close to" $A_i$, so there exists some open set (or ball depending on your exact definition) $O$ that contains $x$ and does not intersect $A_i$. But if it does not intersect $A_i$ it certainly does not intersect $A$. But that would make $x$ not close to $A$, a contradiction. So $x \in A$ must be the case. So $A$ is closed.