3
$\begingroup$

I want to prove that for a closet set $A\subset \mathbb{R}^n$, $\partial A = \partial(\partial A)$. I've already proved that for an arbitrary subset $B$ of $\mathbb{R}^n$, $\partial(\partial B)\subseteq \partial B$ by arriving at the fact that $\partial B\backslash \mbox{interior}(B)\subseteq \partial B$, which points to the fact that the interior of the boundary of a set is not necessarily empty. However, unfortunately, I do not probably understand intuitively how it can be empty. Except that, in $\mathbb{R}^4$, say, the boundary can be a 3-dimensional subset, whose interior does not need to be empty. But then, why should the interior of the boundary of a $\underline{\text{closed}}$ set be necessarily empty? For if we consider the same analogy with $\mathbb{R}^4$, we should also intuitively feel that a boundary can be a 3-dimensional subset, whose interior need not be empty.

Here's also my attempt:

$\partial A= \overline{A}\backslash A^\circ=\overline{\overline{A}\backslash A^\circ}=\overline{A\backslash A^\circ}$.

Now, $(A\backslash A^\circ)^\circ=A\backslash \overline{A^\circ}$. If I can show that the RHS is the empty set then I'll deduce that

$\overline{\overline{A}\backslash A^\circ}\backslash (\overline A\backslash A^\circ)^\circ=\partial (\partial A)=\overline{\overline{A}\backslash A^\circ}=\overline{A}\backslash A^\circ=\partial A$.

I'd appreciate your help.

$\endgroup$
1
$\begingroup$

Let $A \subset \mathbb{R}^N$ and let $a \in A$. We say that $a$ is an interior point of $A$ if there is a positive radius $r > 0$ such that $B(a,r) \subset A$, where $B(a,r)$ is the $N$-dimensional ball centered at $a$ with radius $r$. Notice that the ball you want to consider is $N$-dimensional independently of the dimension of the subset $A$. In particular, if $A$ is an $N-1$-dimensional subset then it cannot contain any $N$-dimensional ball and hence it will have empty interior. Indeed, most of the boundaries that we usually picture in our minds are hypersurfaces and hence have empty interior.

To conclude the proof we need to prove that $\partial A$ has empty interior. This follows from the fact that if $a \in \partial A$, then every neighborhood of $a$ contains a point that is not in $A$. Since $\partial A \subset A$ (recall that by assumption $A$ is closed) we have that every neighborhood of $a$ contains a point that is not in $\partial A$ and hence $a$ cannot be an interior point of $\partial A$.

$\endgroup$
1
$\begingroup$

The boundary $\partial A$ of a closed subset $A \subseteq \mathbb{R}^n$ has empty interior, because any open neighborhood of any point $x \in \partial A$ contains points that are not in $A$ (by definition of boundary), therefore not in $\partial A$ (since $A$ is closed.)

This is certainly not true for arbitrary subsets. The boundary of $\mathbb{Q}^4$ is all of $\mathbb{R}^4$. But then $\mathbb{Q}^4$ is not closed.

$\endgroup$

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.