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$\newcommand{\d}{\operatorname{d}}\newcommand{\b}{\mathcal{B}}\newcommand{\interior}{\operatorname{int}}\newcommand{\ext}{\operatorname{ext}}\newcommand{\bd}{\operatorname{bd}}$In all that follows let $(X,\d)$ be a nonempty metric space.

Some definitions:

The interior of a set $E\subseteq X$, $\interior E$, is the set of all points $x\in E$ such that $\exists r\gt0,\,\b(x,r)\subseteq E$.

The exterior of a set $E\subseteq X$, $\ext E$ is the set of all points $x\in X\setminus E$ such that $\exists r\gt0,\,\b(x,r)\cap E=\varnothing$, i.e. the set of all points interior to $X\setminus E$.

The boundary of a set $E\subseteq X$, $\bd E$ is the set of all points $x$ such that $\forall r\gt0$, $\b(x,r)\cap E\neq\varnothing\neq\b(x,r)\cap X\setminus E$.

A set $E\subseteq X$ is hollow if it has empty interior.

Royden leaves it as an exercise to show that the boundary of any closed subset of $X$ is hollow.

I propose that the boundary of all sets in $X$ is hollow, be they closed, open or neither.

First, I show that: $$\forall E\subseteq X,\,X=\interior E\sqcup\ext E\sqcup\bd E$$

For disjoint union $\sqcup$.

If $x\in\interior E\wedge x\in\ext E$, then $\exists r_1,r_2\gt0:\b(x,r_1)\subseteq E,\,\b(x,r_2)\subseteq X\setminus E$. Since $x\in\b(x,r_1),\,x\in\b(x,r_2)$, this implies $x\in E,\,x\in X\setminus E$, which is impossible. Therefore $\interior E\cap\ext E=\varnothing$.

Now suppose $x\in\bd E\wedge x\in\interior E$. Since $x$ is interior to $E$, there is $r\gt0,\,\b(x,r)\subseteq E$. Since $x$ is a boundary point of $E$, for the same $r$ we have that $\b(x,r)\cap X\setminus E\neq\varnothing$, implying the existence of a $y$ in that intersection, so that $y\in E$ yet also $y\in X\setminus E$, which is impossible. By the same argument the same impossibility holds for $x\in\bd E\wedge x\in\ext E$. Therefore all the sets are disjoint.

If $x\in X$ and $x\notin\interior E\wedge x\notin\ext E$, then there is no $r\gt0$ such that $\b(x,r)\subseteq E\vee\b(x,r)\subseteq X\setminus E$. Then for all $r\gt0$, there is a point in $\b(x,r)$ not in $E$ and a point in $\b(x,r)$ not in $X\setminus E$, implying the intersections are non-empty. Therefore $x\notin\interior E\wedge x\notin\ext E\implies x\in\bd E$.

Now take again an arbitrary $E\subseteq X$. $\bd E$ is clearly closed as a consequence of the above. The exterior of $\bd E$ is then $X\setminus\bd E$, and again by the above we have that $\interior\bd E\sqcup\bd\bd E=\bd E$. Every point $x\in\bd E$ satisfies $\forall r\gt0,\,\b(x,r)\cap X\setminus\bd E\neq\varnothing$, and of course every $x\in\bd E$ satisfies $\forall r\gt0,\,\b(x,r)\cap\bd E\neq\varnothing$. Then $\bd\bd E=\bd E$, from which it follows by the disjoint nature of the union that $\interior\bd E=\varnothing$, that is, $\bd E$ is hollow.

Are my proof and the result correct? I've tried to look it up but have seen no reference to this.

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    $\begingroup$ Isn't $\mathbb Q \subset \mathbb R$ a counterexample? Or am I misreading the definitions? $\endgroup$
    – Ravi Fernando
    Commented Jan 17, 2022 at 19:08
  • $\begingroup$ @RaviFernando What do you propose to be the boundary of $\Bbb Q$? $\endgroup$
    – FShrike
    Commented Jan 17, 2022 at 19:14
  • $\begingroup$ All of $\mathbb R$. (Certainly $\mathbb Q$ has no interior or exterior.) $\endgroup$
    – Ravi Fernando
    Commented Jan 17, 2022 at 19:15
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    $\begingroup$ Suppose $E=\mathbb{Q}\cap [0;1]\in\mathbb{R}$, then it seems $bd(E)=[0;1]$ whereas $bd( bd(E))=\{0,1\}$ ? $\endgroup$
    – QuantumPotatoïd
    Commented Jan 17, 2022 at 20:08
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    $\begingroup$ @Cretin2 That's right. It would seem that $\Bbb Q$ is pathological to my intuition that a set cannot have thick boundary $\endgroup$
    – FShrike
    Commented Jan 17, 2022 at 20:11

1 Answer 1

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$\newcommand{\d}{\operatorname{d}}\newcommand{\b}{\mathcal{B}}\newcommand{\interior}{\operatorname{int}}\newcommand{\ext}{\operatorname{ext}}\newcommand{\bd}{\operatorname{bd}}$

Without assuming that $E$ is closed, the boundary of $E$ might have non-empty interior. For example, let $X = \mathbb R$ and $E = \mathbb Q$. Then \begin{align} \interior E &= \emptyset,\\ \ext E &= \emptyset,\\ \bd E &= \mathbb R. \end{align} So the boundary has non-empty interior.

Your proof failed at

Every point $x\in\bd E$ satisfies $\forall r\gt0,\,\b(x,r)\cap X\setminus\bd E\neq\varnothing$,

The last $\bd E$ should be replaced by $E$ by definition of $\bd E$.

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    $\begingroup$ I agree with your first objection but not your second: $x$ itself lies in $\mathcal B(x, r) \cap \mathrm{bd} E$. $\endgroup$
    – Ravi Fernando
    Commented Jan 17, 2022 at 19:19
  • $\begingroup$ I have deleted that (I will check if the OP had another mistake.... now), thank @RaviFernando $\endgroup$
    – Arctic Char
    Commented Jan 17, 2022 at 19:21
  • $\begingroup$ Aha. I made the assumption that neither the interior nor the exterior of $E$ is empty, that is, I assumed $E$ is not simultaneously hollow and dense as is the case in the weird situation of $\Bbb Q$. Does the result hold if we add that assumption? I am finding it difficult to conceptualise the idea that a set might have thick boundary. $\endgroup$
    – FShrike
    Commented Jan 17, 2022 at 19:27
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    $\begingroup$ No unfortunately. One can take $E = (1,2) \cup ((-2, -1)\cap \mathbb Q)$. Then $E$ non-empty interior and exterior, but $bd E = \{ 1,2\} \cup [-2,-1]$. @FShrike $\endgroup$
    – Arctic Char
    Commented Jan 17, 2022 at 19:31

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