I already looked at about 3 problems on here (listed below), but they all seemed to use metric spaces. I was trying to understand this proof (from armstrong's basic topology), which just works in a general topological space. Basically, Armstrong shows for A contained in a topological space X, cl(A) is closed because X – cl(A) is open. He first shows if we take an element x in X – cl(A) , we can find an open neighborhood U around it that contains no elements of A (I think this is clear by definition of the complement). Then he says U can't contain any limit points since it's open. I think I understand everything expect that: Why would being a neighborhood of every point mean U couldn't contain any limit/accumulation points of A:
These are articles I referenced, but I couldn't generalize it not in a metric space:
Closure of a set is closed proof
Prove the closure is closed and is contained in every closed set
Best Answer
One definition of the closure is as the intersection of all closed sets containing $A$. Or, the smallest closed set containing $A$. Hence closed.
As to the part you didn't understand: a limit point of $A$ can't have a neighborhood disjoint from $A$ (by definition of limit point).