Metric Space - Accumulation Let $(X,d)$ be a metric space. Let $A \subset X$ and $c \in X$. $c$ is called an accumulation point of $A$ if for every $\delta > 0$ there exists $a \in A$ such that $0 < d(a,c) < \delta$. The set of accumulation points of a set $A$ is denoted by $A'$. For example if $\Bbb R$ is endowed with the usual metric, then $\Bbb Q' = \Bbb R$, $\Bbb N'= \varnothing$, $(a,b)' = [a,b]$. Let $A$, $B$, $C$, $D$ be subsets of $X$. Prove the following statements.
(a) $C \subset D \Longrightarrow C' \subset D'$
(b) $(A \cup B)' = A' \cup B'$
(c) $\overline{A} = A \cup A'$
(d) $A$ is closed if and only if $A' \subset A$
(e) If $B$ is finite, then $B' = \varnothing$
(f) If $B$ is a finite subset of $A$, then $A' = (A \setminus B)'$. Note that $A = B \cup (A \setminus B)$
(g) Let $(x_n)$ be a sequence in $X$. If $A = \{x_n : n \in \Bbb N\}$ and $a \in A'$, then $(x_n)$ has a subsequence converging to $a$. (Use induction and (f) to construct a strictly increasing sequence $(k_n)$ of integers such that $0<d(x_{k_n},a) < 1/n$)
 A: HINTS: Parts (a), (c), and (e) are very easy: each of them is just a matter of using the definition. For instance, to prove (a), you must show that if $x\in C'$, then $x\in D\,'$. Suppose that $x\in C'$; then by definition for each $\delta>0$ there is some $c\in C$ such that $0<d(x,c)<\delta$. But $C\subseteq D$, so $c\in D$. Thus, for each $\delta>0$ there is some $c\in D$ such that $0<d(x,c)<\delta$, which by definition means that $x\in D\,'$.
For (c) you need to recall that $x\in\overline A$ if and only if for each $\delta>0$ there is an $a\in A$ such that $d(x,a)<\delta$. Suppose that $x\in A\cup A'$, and let $\delta>0$. If $x\in A$, you can take $a$ to be $x$ itself, since $d(x,x)=0<\delta$, and if $x\in A'$, then there is an $a\in A$ such that $0<d(x,a)<\delta$, which certainly implies that $d(x,a)<\delta$!
For (e), let $B=\{b_1,\dots,b_n\}$ be a finite set, and let $x\in X$. What happens if you set $$\delta=\min\{d(x,b_k):k=1,\dots,n\text{ and }x\ne b_k\}\;?$$ Is $\delta>0$? Can you find a $b_k\in B$ such that $0<d(x,b_k)<\delta$?
Once you have (c), (d) is very easy; just remember that if $A$ is closed, then $\overline A=A$.
Half of (b) follows immediately from (a): $A\subseteq A\cup B$, so $A'\subseteq(A\cup B)'$, and similarly $B\,'\subseteq(A\cup B)'$, so $A'\cup B\,'\subseteq(A\cup B)'$. Thus, it only remains to show that $(A\cup B)'\subseteq A'\cup B\,'$. To prove this, let $x\in(A\cup B)'$, and show that $x\in A'\cup B\,'$, i.e., that either $x\in A'$ or $x\in B\,'$. The easiest way is probably to suppose that $x\notin A'$ and $x\notin B\,'$ and derive a contradiction by showing that $x\notin(A\cup B)'$ after all. Note: The definition of $A'$ immediately tells you that if $x\notin A'$, then there is some $\delta>0$ such that no $a\in A$ satifies the inequality $0<d(x,a)<\delta$: every point of $A$ is either equal to $x$ or at least $\delta$ distant from $x$.
The second sentence in (f) is a hint: apply (b) to $B\cup(A\setminus B)$. You’ll also want to use (e) here. 
That’s quite a bit to work on, so I’m going to stop here. Part (g) is harder than the rest, but that’s why it comes with a very extensive hint; perhaps you’ll be ready to tackle it on your own after you’ve dealt with the first six parts.
