Prove that the set $A’ = \{x \in \mathbb{R}^k \mid x \in \partial (A/\{x\})\} $ is closed Let $A$ be a set in $\mathbb{R}^k$.
Prove that the following set:
$$
A’ = \{x \in \mathbb{R}^k \mid x \in \partial (A/\{x\})\}
$$
is a closed set.

What i tried:
Lets try to look at every convergent sequence in $A’$, we will show that it converge to an element $\in A’$, therefore the set is closed. (Its not working...)
The problem is that i dont know how to prove it:
Assume that $\{x_n\}$ is a convergent set in $A’$, now lets assume that it converges to a point $L \notin A’$. Now i dont see any problem that the points will become closer to a point not in $A’$...
Another way is to look at the complement set, if i prove that the complement is open, i proved what is neede, yet here it is very confusing, it satisfies:
$$
\overline{A} = \{ x \in \mathbb{R}^k \mid x \notin \partial (A/\{x\}) \}
$$
This is very confusing and i dont know how to decide if its open or close set, can someone give me a hint?
Thank you.
 A: Let's take the complement of $A'$ as you defined it: $x \notin A'$ iff $x \notin \partial (A\setminus\{x\})$, which means by definition that there is some ball $B(x,r)$ around $x$ that does not intersect $A \setminus \{x\}$ or the complement of $A\setminus\{x\}$. But $x \in B(x,r)$ always and $x$ is in the complement of $A\setminus\{x\}$ so that the second options does not hold.
So $B(x,r) \cap (A\setminus\{x\}) = \emptyset$ holds for some $r>0$.
Now, I claim that in fact $B(x,r)$ is a subset of the complement of $A'$ so that $x$ is an interior point of $\Bbb R^k \setminus A'$ and so $A'$ is closed, as its complement is open.
To see the claim, it suffices to show that any $y \in B(x,r)$ that is not equal to $x$ is not in $A'$. (for $x$ we already know this).
So let $y \neq x$ be given in $B(x,r)$. Then let $r' = \min(d(x,y), r-d(x,y)) > 0$
Then the choice of $r'$ ensures that $x \notin B(y,r')$ and $B(y, r') \subseteq B(x,r)$ and so $B(y,r') \cap (A\setminus \{y\}) = \emptyset$. So This ball witnesses that $y \notin \partial (A\setminus\{y\})$ and hence $y \notin A'$. QED.
