Is this true?: converging sequence question Let X be a metric space, and let A ⊂ X. Suppose that {pn} is a sequence in A
which converges to some point p ∈ X. True or false: i) p ∈ A′ (limit points of A)
(ii) p ∈ closure(A)
These are both true, right? And if (i) is true isn't (ii) always true (because closure(A)=A union A')
 A: Suppose $p_n = p$ for all $n$. Then $p_n$ converges to $p$, but $p$ might not be a limit point of $A$. It might be an isolated point of $A$.
To give a concrete example, suppose that $A = \{0\} \cup [1,2]$ and $p_n = 0$ for all $n$. Then $p_n \rightarrow 0$ but $0$ is not a limit point of $A$.
What you can say for sure is that if $p_n \in A$ and $p_n \rightarrow p$, then $p$ is either an isolated point of $A$ or a limit point of $A$. In either case, $p \in \text{closure}(A)$ since $\text{closure}(A) = A \cup A'$ and every isolated point of $A$ is contained in $A$.
Referring to your problem statement, (i) implies (ii) but (ii) does not imply (i). And $p_n \rightarrow p$ implies (ii) but not necssarily (i).
Note that a point $x \in X$ is a limit point of $A$ if and only if every neighborhood of $x$ contains a point of $A$ distinct from $x$. This in fact is equivalent to the condition that every neighborhood of $x$ contains infinitely many elements of $A$. Another equivalent condition is that there exists a sequence $x_n \in A$ of distinct values (no repeats) which converges to $x$.
Given any set $A$, every element of $A$ is either an isolated point of $A$ or a limit point of $A$. Also, $A$ contains all of its isolated points but it need not contain all of its limit points.
