Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $0<d(x_{k_n},a) < 1/n$)

share|cite|improve this question
What do you think? What have you tried? How can you apply the definitions you need to tackle these problems? – amWhy Dec 2 '12 at 22:33
Actually, I am a physics student and I joined a class in Math dept. because I really want to learn Functional Analysis. Because of that I don't have basic information, I could not find a solution to this question myself. I will be appreciated if you can help me. – Ersin Dec 2 '12 at 23:03

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.