# isolated point in boundary

Let $A$ be a subset of a metric space $X$ and let $x_o$ be an isolated point of $A$. Show $x_o\in\partial A$ iff $x_o\in Acc(A^{c})$.

My attempt:

($\rightarrow$) Let $x_o\in\partial A$. Then, $B_\epsilon(x_o)\bigcap A \neq \emptyset$ and $B_\epsilon(x_o)\bigcap A^{c} \neq \emptyset$, $\forall\epsilon>0$. Using the latter fact, we see $x_o\in Acc(A^{c})$.

($\leftarrow$) Let $x_o\in Acc(A^{c})$. Then, $\forall\epsilon>0,B_\epsilon(x_o)\bigcap A^{c} \neq \emptyset$ Since $x_o$ is an isolated pt. of A, $\exists\epsilon>0$ s.t. $B_\epsilon(x_o)\bigcap A={x_o}\neq\emptyset$. Thus $x_o\in\partial A$.

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The argument is okay, though you’ve worked a little harder than necessary in the second part ($\leftarrow$). It isn’t important that $x_0$ is an isolated point of $A$: all you need is that $x_0\in A$.
In both directions the fact that $x_0\in A$ automatically tells you that $B_\epsilon(x_0)\cap A\ne\varnothing$ for every $\epsilon>0$, so you’re really only concerned with $B_\epsilon(x_0)\cap A^c$.