Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

A boundary point $z∈A$ may not be an accumulation point.

proof: If $z$ is an isolated point of $A$ (i.e. there is a ball $B(z,r)$ such that $B(z,r)\cap A=\{z\}$) then $z$ is a boundary point but not an accumulation point.

But if $z$ is a boundary point and $B(z,r)\cap A=\{z\}$, then should not it follow that $r=0$? Otherwise there must be other point in the intersection. But in that case $z$ should also be an accumulation point because if we take $r>0$, $B(z,r)\cap A$ will contain a point $k$, $k\in A$ and $k\neq z$.

In that case $z$ is always an accumulation point. What is wrong with my thought could you please explain it to me?

share|cite|improve this question
Amadeus, I think your reasoning is correct, actually, and there can be isolated boundary points of A (hence not accumulation points). Perhaps the Venn diagram in Section 3 of this might help you understand the categories better? – Conan Wong Dec 4 '12 at 19:22
Let $A=\{0\} \subset \mathbb{R}$. Then $B(0,r) \cap A = \{0\}$ for all $r>0$. – copper.hat Dec 4 '12 at 19:26
up vote 2 down vote accepted

Consider the set $\mathbb{Z}$ of integers in the real line $\mathbb{R}$. Clearly for each $n \in \mathbb{Z}$ there is an $r > 0$ such that $B ( n , r ) \cap \mathbb{Z} = \{ n )$: just take any $0 < r < 1$.

Note that for $x$ to be a boundary point of a set $A$ for each $r > 0$ there must be points $z_0 , z_1 \in B ( x , r )$ such that $z_0 \in A$ and $z_1 \notin A$. If $x$ itself is an element of $A$, then clearly for each $r > 0$ there is a $z_0 \in B ( x , r )$ such that $z_0 \in A$: just take $z_0 = x$! So for a point $x \in A$ to belong to the boundary of $A$ we just need that for each $r > 0$ there is a $z_1 \in B ( x,r )$ with $z_1 \notin A$; i.e., we need $x \in \overline{ X \setminus A }$. This really has nothing to do with $x$ being an accumulation point of $A$ (as the example I give above shows).

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.