Thread: metric space

Hello

Originally Posted by patricia-donnelly

Let (X,d) be a metric space and say A is a subset of X. If x is an accumulation point of A, prove that every r-neighbourhood of x actually contains an infinite number of distinct points of A (where r>0).

Suppose is an accumulation point of a set . Then there exists distinct such that . Let then by definition of convergence it means for . Thus, are distinct elements which lie in and hence there are infinitely many.

Using this, prove that any finite subset of X is closed.

A set is closed if and only if it contains all its accumulation points. A finite set clearly has no accumulation points, hence the set is empty. And it contains the empty set.

You've been a great help. Thank you very much