2
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ 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. $\endgroup$ – Greg Martin Mar 6 at 8:07
  • 1
    $\begingroup$ Note that $A' = \{x \in \Bbb R^k \mid \forall r>0: B(x,r) \cap (A\setminus \{x\} \neq \emptyset\}$. $\endgroup$ – Henno Brandsma Mar 6 at 8:44
  • $\begingroup$ 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! $\endgroup$ – Gamow Drop Mar 6 at 9:23
  • 1
    $\begingroup$ $\partial A$ can be $\Bbb R^k$ and not any subset of it is closed... $\endgroup$ – Henno Brandsma Mar 6 at 9:26
  • 1
    $\begingroup$ @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. $\endgroup$ – Greg Martin Mar 6 at 18:55
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.