# The boundary of any set is hollow

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

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

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

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.

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 $$\bd E$$.
• I agree with your first objection but not your second: $x$ itself lies in $\mathcal B(x, r) \cap \mathrm{bd} E$. Commented Jan 17, 2022 at 19:19
• 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. Commented Jan 17, 2022 at 19:27
• 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 Commented Jan 17, 2022 at 19:31