Show that $\operatorname{Cl}(A)\setminus A$ consists of entirely of accumulation points of $A$

Question

Let $(M,d)$ be a metric space and $A \subset M$. Show that $\operatorname{Cl}(A)\setminus A$ consists of entirely of accumulation points of $A$, where $\operatorname{Cl}(A)$ is the closure of $A$.

My attempt:

Suppose $x \in \operatorname{Cl}(A)\setminus A$ is not an accumulation point, $\exists$ $\epsilon > 0$, s.t. $B(x, \epsilon)\cap A$ is empty, i.e. $B(x, \epsilon) \subset A^c$.

I do not know where to go from here; I was hoping to show that either $\operatorname{Cl}(A)$ could be even smaller or that $x \notin \operatorname{Cl}(A)\setminus A$ (a contradiction either way).

I appreciate any help. Thank you very much!

Answer 1

Saying that $x$ is not an accumulation point of the set $M$ means that there is $\varepsilon>0$ such that $B(x,\varepsilon)\cap M\subseteq\{x\}$.

You're making a mistake in negating the definition of accumulation point:

$x$ is an accumulation point of $M$ if (and only if), for every $\varepsilon>0$, $B(x,\varepsilon)\cap M$ contains a point distinct from $x$.

It's easier if you do a direct proof. Suppose $x\in\operatorname{Cl}(A)\setminus A$ and let $\varepsilon>0$. Then $B(x,\varepsilon)\cap A\ne\emptyset$ by definition of closure. If $y\in B(x,\varepsilon)\cap A\ne\emptyset$, then $y\ne x$, because $y\in A$ and $x\notin A$.

But your idea, with the fix, still works: suppose $x\notin A$ is not an accumulation point of $A$; then there exists $\varepsilon>0$ such that $B(x,\varepsilon)\cap A\subseteq\{x\}$. Since $x\notin A$, we conclude that $B(x,\varepsilon)\cap A=\emptyset$ and therefore $x\notin\operatorname{Cl}(A)$.

Answer 2

Suppose $x \in \operatorname{Cl}(A) \setminus A$. Then every $B(x,r)$ intersects $A$ (as $x \in \operatorname{Cl}(A)$) but $B(x,r)$ must intersect $A$ in another point than $x$ (because $x \notin A$, trivially). Ergo, $x \in A'$.

Answer 3

First $\operatorname{cl}A=A\cup A'$.

Thus $\operatorname{cl}A\setminus A=(A\cup A')\setminus A=A'\setminus A\subset A'$.