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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?

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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? en.wikipedia.org/wiki/Boundary_%28topology%29#Examples –  Conan Wong Dec 4 '12 at 19:22
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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

1 Answer 1

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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).

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