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?