Prove that $ x \in \overline{A} $ iff there is a sequence of points of $ A $ that converges to $ x $

Question

Question from Mendelson's Intro to Topology chapter 2: Let $ A $ be a subset of a metric space $ (X, d) $. Let $ A' $ be the set of limit points of $ A $ and $ A^{i} $ be the set of isolated points of $ A $. Denote $ \overline{A} = A' \cup A^{i} $. Prove that:

a) $ A' \cap A^{i} = \emptyset $ and $ A \subset A' \cup A^{i} $.

b) $ x \in \overline{A} $ iff there is a sequence of points of $ A $ that converges to $ x $.

c) If $ F $ is a closed set such that $ A \subset F $, then $ \overline{A} \subset F $. A set is closed if its complement is open. A set is open if it is a neighborhood of each of its point.

d) $ \overline{A} $ is the intersection of all such closed sets $ F $.

I have successfully proved part a) but currently stuck on b). Can anyone drop a hint or suggestion on how to approach?

Answer 1

One direction: You need to show that if $x \in \overline{A}$, there is a sequence of points in $A$ converging to $x$. $x$ must be in either $A'$ or $A^i$, so treat the two cases separately, using the definition directly for the first case and notice that the second case isn't too difficult. On the other hand, if there is a sequence of points in $A$ converging to $x$, consider the cases where $x$ is or isn't in $A$, and show it is either in $A^i$ or $A'$.