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I am reading D. J. H. Garling's a Course in Mathematical Analysis, and I came across the following problem:

12.3.7. Show that the interior of the boundary of a subset of a metric space is empty.

I've been working on this for a while, but I can't seem to figure it out. For one, the statement is clear, when talking about open, or closed sets. But I have struggle with the general case. I've tried the following:

Let be given a metric space $(X, d)$, and $A \subseteq X$. Suppose that the $x \in (\partial A)^{o}$. Then, $\exists \> \varepsilon > 0$ st. the open ball $B_{\epsilon}(x) \subseteq \partial A$. Here I should use the fact that every point in the ball is in the

Alternatively, I tired just using the fact that, if $x \in \partial A$, then $\forall \varepsilon > 0 \> \exists y \in A, z \in C(A)$, st. $y, z \in B_{\varepsilon}(x)$. This yielded simlarly little success.

I am looking for a realitvely pure argument, as I found some complicated ones online, but I believe there should be a clean solution. Thanks!

blomp
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    $x \in \partial A\Rightarrow x\notin A^{\circ}$. – Piquito Jul 08 '23 at 12:57
  • Thanks for the reply. I think the issue was, as it is pointed out in the accepted answer, that the problem was ill-posed. The corrected version is easy to prove, though, and I have no problems with that. – blomp Jul 09 '23 at 10:12

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You cannot prove it, since it is false. For instance, take $\Bbb Q$, as a subset of $\Bbb R$, endowed with its usual topology. Then $\partial\Bbb Q=\Bbb R$, whose interior is $\Bbb R$ itself.

José Carlos Santos
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