show that the set $ A'=\left\{ x:x\in\partial(A\setminus\{x\}\right\} $is closed I'm studying Calculus 3, and our teacher tends to give us extremely difficult questions.
one of those questions was this:
"let $A \subseteq \Bbb R^k $. show that the set $ A' = \{ x \in \Bbb R^k\mid x \in \partial(A \setminus \{ x \})\}$ is closed".
I've tried to show using sequence characterization or showing that $(A')^c$ is open, but I don't have a clue of how to start, or even if I'm in the right direction.
I could really use some help or a guiding hand with this.
 A: I'll use $d$ for the metric, and work in any metric space instead of $\Bbb R^n$.
Suppose that $x \notin A'$. This means that $x \notin \partial(A\setminus \{x\})$, so either there is an open ball $B(x,r_x), r_x>0$ such that $B(x,r_x) \subseteq A\setminus \{x\}$ or such that $B(x, r_x) \cap (A\setminus \{x\}) = \emptyset$; the former is impossible (contradicted by $x$) so the latter is the case.
My claim is that $B(x,r_x) \cap A' = \emptyset$, showing that $A'$ is closed.
Let $y \in B(x,r_x)$. If $y=x$, then we already know that $y \notin A'$.
So assume $y \neq x$. We can find $r_y>0$ so that $B(y, r_y) \subseteq B(x,r_x)$ and moreover $d(y,x) \ge r_y$ so that $x \notin B(y, r_y)$ as well. Now note that $B(y, r_y) \cap A\setminus \{y\} = \emptyset$ (being a subset of $B(x,r_x)$ avoids all points of $A$, except maybe $x$, but choosing $x$ outside of that ball takes care of that), and this means that $y \notin \partial(A\setminus \{y\})$ or $y \notin A'$, as required.
A: Since you are looking at the boundary of $B = A-{x}$, the point $x$ is the point not in $B$ for any ball around $x$. So if $x$ is in the boundary of $B$ then $x$ is a limit point of $A$ since in the definition of limit point, every neighbourhood of the point $x$ in question must contain a point of the set $A$ other than $x$ itself. The set of all limit points of a set $S$ is called the derived set of $S$.
So based on how you have defined the boundary in your class the set you are analyzing $\{x\in \mathbb{R}^{k}| x\in A\setminus{\{x\}}$ is called the derived set of $A$.
So if you can use the theorem:
If $X$ is a $T1$ space, the derived set of every subset of $X$ is closed in $X$.
Then the derived set of $A$ is closed. If not we can elaborate that proof a bit more.
Let's call the derived set of $A$, $A'$.
Let q be a limit point of A′. If it is proved that $q \in A′$, then the proof is done.
Let $G_{q}$ be the open set containing q. Since q is a limit point of A′,$G_{q}$ contains at least one point $r \in A′$ different from q. But $G_{q}$ is an open set containing $r \in A′$; hence $G_{q}$ contains infinitely many points of A.
So there exist $a \ in A$ such that $a\neq q,a\neq r$ and $a\in G_{q}$. That is,each open set containing q contains infinitely many points of A. Hence $q\in A′$.
A: I think that defining boundaries without closures and interiors is a kind of self-torturing. However let me clarify the correct proof of Mr.Henno Brandsma!
Claim $1$: $X/A-\left\{ a\right\}=A^{c}\bigcup\left\{a \right\}$
Proof: If $x\notin A-\left\{ a\right\}$ then i)if $a\notin A$ then $A-\left\{ a\right\}=A$ hence $x\notin\,A$, i.e. $x\in\,A^{c}$ ii) if $a\in$$A $ and assume $x\notin\,A^{c}\bigcup\left\{a \right\}$ then $x\,\in A$ and $x\neq\,a$ which gives $x\,\in\,A-\left\{a \right\}$, contradiction. Thus $x\,\in\,A^{c}\bigcup \left\{a \right\}$.
Now, assume $x\,\in\,A^{c}\bigcup \left\{a \right\}$. If $x\,\notin\,A$ clearly  $x\notin A-\left\{ a\right\}$ which is a smaller set. If $x=a$, then obviously  $\,x=a\notin A-\left\{ a\right\}$. So Claim $1$ is proved.
Now in the proof of Mr.Brandsma we have $B(x,r_{x})\bigcap A-\left\{x \right\}=\varnothing$. According to Claim $1$ this implies $B(x,r_{x})\subseteq A^{c}\bigcup \left\{ x\right\}$.
Now intersect both sides with  $B(y,r_{y})$. We obtain
$B(y,r_{y})\subseteqq A^{c}\bigcap B(y,r_{y})$ (because $x\notin\,B(y,r_{y})$) which implies $B(y,r_{y})=A^{c}\bigcap B(y,r_{y})$ , hence $B(y,r_{y})\subseteqq A^{c}$ and clearly
$B(y,r_{y})\subseteqq A^{c}\bigcup\,\left\{y \right\}$ which gives $B(y,r_{y})$$\bigcap\,A-\left\{y \right\}$= $\varnothing $ and hence $y\notin A'$ as it was required!!
