Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. How do I prove that $A'$ is closed?

Question

I was asked this simple question a short while ago.

Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. Prove that $A'$ is closed.

I think I was able to explain it, but I am not sure what I claim is true. (The following is not a formal solution because I am not sure the reasoning is correct) My explanation goes as follows:

Let $x\in A'$. Say $x$ is in the interior of $A$, then $x\notin\partial(A\setminus\{x\})$ because there exists a ball $B(x;r)\subseteq A$ by definition. By a smilar logic if x is in the exterior of $A$, I claim that $x\notin\partial(A\setminus\{x\})$. Thus $x\in\partial A$ (that is $A'\subseteq\partial A$). From here I am quite sure it is not too hard to explain that $A'$ is closed.

I doubt my reasoning, but I am not sure what have I missed. Care to shed some light over the matter?

Answer 1

Let $p \notin A'$. This means that $\exists r>0: B(p,r) \cap A \subseteq \{p\}$. If in fact $B(p,r) \cap A = \emptyset$, for any $x \in B(p,r)$ we have a ball $B(x,r') \subseteq B(p,r)$ (open balls are open sets) and this witnesses also that $B(x,r') \cap A= \emptyset$ and so $x \notin A'$. So $B(p,r) \subseteq \Bbb R^k \setminus A'$. If on the other hand, $B(p,r) \cap A= \{p\}$, we also have $B(p,r) \subseteq \Bbb R^k \setminus A'$: if $x \in B(p,r)$ and $x \neq p$ (WLOG) then there is a ball $B(x,r') \subseteq B(p,r)$ so that $p \notin B(x,r')$. Then $B(x,r') \cap A = \emptyset$ and so $x \notin A'$, as required. So any $p \in \Bbb R^k \setminus A'$ is an interior point of it, so the complement of $A'$ is open, and $A'$ is closed.

Answer 2

Suppose that $(x_n)_{n\in\Bbb N}$ is a sequence of elements of $A'$ which converges to some $x\in\Bbb R^k$; I will prove that $x\in A'$. Take $r>0$ and consider the ball $B(x;r)$. It contains some $x_n$. Since $x_n\in A'$, $x_n\in\partial(A\setminus\{x\})$. So, every ball $B(x_n;r')$ contains elements of $A$. So, take $r'$ so small that $B(x_n;r')\subset B(x;r)\setminus\{x\}$. Then $B(x_n;r')$ contains elements of $A\setminus\{x\}$. This proves that $x\in A'$. Therefore, since $A'$ contains the limit of any convergent sequence of its elements, it is a closed set.

Answer 3

I believe you get a better insight if you forget about $\mathbb{R}^k$ and balls. Let's work in a (Hausdorff) topological space $X$, which your case is a specialization of.

What are the points $x$ such that $x\in\partial(A\setminus\{x\})$?

First of all they don't need to belong to $A$. For instance, with $A=(0,1)\subseteq\mathbb{R}$, $0$ belongs to the set $A'$.

Let's see: $x\in\partial(A\setminus\{x\})$ means that every neighborhood of $x$ intersects both $A\setminus\{x\}$ and its complement. However, $x$ surely belongs to the complement of $A\setminus\{x\}$, so the condition just reads $$ \textit{every neighborhood of $x$ intersects $A\setminus\{x\}$} $$ which is the standard definition of limit point.

Now it's easier, isn't it? You need to show that if $y$ has the property that each of its open neighborhoods intersects $A'$, then $y\in A'$.

Take such an element $y$ and an open neighborhood $V$ of $y$. Then there exists $x\in A'$ such that $x\in V$. Since the space $X$ is Hausdorff, there exists a neighborhood $U$ of $x$ such that $U\subseteq V$ and $y\notin U$.

Since $U$ is a neighborhood of $x$, there exists a point $z\in A$ (different from $x$, but it's irrelevant) such that $z\in U$. Now necessarily $z\ne y$, but $z\in V$ and we're done.