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In general, what is the relation between the set of all accumulation points of a a subset of a metric space, say $A$, and its closure ?

For any $a \in A$, if $a \in IntA$, then $a \in Accum(A)$. However, I could not derive the same thing for the boundary of $A$

Edit:

These are the definitions that I'm using:

$$\{\text{accumulation points of }A\} = \{ a\in E | \forall \epsilon > 0 \quad (B(a, \epsilon) - \{a\} )\cap A \not = \emptyset\}$$

$$\partial(A) = \{ a \in E | \forall \epsilon >0 , B(a, \epsilon) \cap A \not = \emptyset \quad and \quad B(a, \epsilon) \cap A^c \not = \emptyset \}$$

$$\{\text{accumulation points of }A\} = int(A) \cup \partial (A)$$

Our
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1 Answers1

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What you're after is the derived set of the set $A$, which is denoted by $A'$. In general $A'\subset\overline A$. When we have $A'=A$, we say that $A$ is a perfect set. On the other hand, there are cases in which $A\varsupsetneq A'\varsupsetneq A''\varsupsetneq\cdots$. For instance, if$$A=\left\{\frac1n+\frac1m\,\middle|\,m,n\in\mathbb{N}\right \},$$then$$A'=\left\{\frac1n\,\middle|\,n\in\mathbb{N}\right\}\cup\{0\}$$and $A''=\{0\}$ (and, of course, after this all derived sets are equal to $\{0\}$)..

José Carlos Santos
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  • Actually, I was going to ask "How to prove that p is an accumulation point of A iff $\exists (x_n) \in A$ s.t $\lim x_n = p$" after this question.Therefore, I am going to ask you to prove it, or provide a reference for the proof of it. – Our Nov 11 '17 at 07:43
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    @onurcanbektas Please define “accumulation point”. – José Carlos Santos Nov 11 '17 at 07:50
  • Plus, consider $A = { 1/n | n\in \mathbb{N}}$. The only accumulation point of $A$ is $0$, whereas $1$ is in the boundary of $A$ ? – Our Nov 11 '17 at 07:50
  • @onurcanbektas Again, please define “accumulation point”. – José Carlos Santos Nov 11 '17 at 07:51
  • ${\text{accumulation points of }A} = { a\in E | \forall \epsilon > 0 \quad (B(a, \epsilon) - {a} )\cap A \not = \emptyset}$. – Our Nov 11 '17 at 07:51
  • @onurcanbektas Then$${\text{accumulation points of }A}=\overline A$$by definition. – José Carlos Santos Nov 11 '17 at 07:54
  • I have forgot one term in the definition, i.e "- ${a }$", but edited now. – Our Nov 11 '17 at 07:54