Prove that, if S is open, each of its points is a point of accumulation of S.
How can I prove this??
It's clearly untrue in arbitrary spaces, but since this is an advanced calculus book I'm going to take a wild guess and assume you meant and open subset of . If so then this is true if is non-empty. Let and any neighborhood of it, then since is open we know there exists a neighborhood of such that . Note then that is a neighborhood of contained in . Choose some open ball and use the uncountability of open balls in euclidean space to find a point different from in .