# 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?

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?

• It's not true that $A'=\partial A$ in general, and not every subset of a closed set is closed, so there's definitely still an argument to be made. It seems actually that your observations might more directly prove that the complement of $A'$, namely $\big\{ x\in\Bbb R^k\colon x\notin\partial(A\setminus\{x\}) \big\}$, is open. – Greg Martin Mar 6 at 8:07
• Note that $A' = \{x \in \Bbb R^k \mid \forall r>0: B(x,r) \cap (A\setminus \{x\} \neq \emptyset\}$. – Henno Brandsma Mar 6 at 8:44
• I think I get it! By my claims I can show that $\mathbb{R}^k{\setminus}A'$ is open, which is very similar to what @HennoBrandsma showed in his proof and pointed in his comment. Thank you for your speedy replies everyone! – Gamow Drop Mar 6 at 9:23
• $\partial A$ can be $\Bbb R^k$ and not any subset of it is closed... – Henno Brandsma Mar 6 at 9:26
• @GamowDrop: are you thinking that any subset of a closed set must be closed? That's definitely not true. For a concrete example, suppose $A$ is the unit disk in $\Bbb R^2$, so that $\partial A$ is the unit circle. The subset of $\partial A$ consisting of all points strictly about the $x$-axis is not closed. Neither is the set of points in $\partial A$ that have both coordinates rational. – Greg Martin Mar 6 at 18:55

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.

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.

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.