-1

Let $(X,d)$ be an arbitrary metric space, with $X$ being the set, and $d$ the metric. Let $S$ be a subset of $X$. Let $bd(S)$ be the boundary of $S$, in metric space $(X,d)$.

I want to show that if the interior of $bd(S)$ is equal to $S$, then $S$ is the empty set.

I encountered this question as part of Exercise C.9 in Efe Ok's "Real Analysis with Economic Applications," which I am reading for self-study and review, so this is not for an assignment. The exercise has several other parts, but I figured out those, unlike this one.

What I have so far hasn't helped me much. My plan was to assume that $x$ is an arbitrary element in the interior of $bd(S)$, and then derive that $x$ cannot exist. The "assume" part of that plan was fine, but the "derive" not so much. Can anyone who finds this question easy give me a little push to get me going? I'd love a hint even better than a full solution.

Arnaud Mortier
  • 27,276
Jon E.
  • 3

1 Answers1

0

Suppose otherwise. Then there is some $x\in S$. Therefore, $x$ belongs to the interior of $\operatorname{bd}S$. In other words, there is some $r>0$ such that $B(x,r)\subset\operatorname{bd}S$. Since $B(x,r)$ is an open set, it is therefore a subset of the interior of $\operatorname{bd}S$, that is, it is a subset of $S$. But then $B(x,r)\subset\mathring S$. In particular, $x\in\mathring S$. This is impossible, because $x\in\operatorname{bd}S=\overline S\setminus\mathring S$.

José Carlos Santos
  • 427,504
  • I had gone as far as B(x,r) is a subset of the interior of bd(S). But then... why can we infer that it follows that B(x,r) is a subset of S? That's the step I got stuck at: if B(x,r) is contained in S, then sure, x is an interior point and we are done. But the interior of bd(S) need not be contained in S, need it? – Jon E. Apr 20 '18 at 20:50
  • @JonE. Since $B(x,r)$ is a subset of the interior of $\operatorname{bd}S$ and since the interior of $\operatorname{bd}S$ is equal to $S$, $B(x,r)$ is a subset of $S$. – José Carlos Santos Apr 20 '18 at 20:52
  • Crystal clear now. I had lost track of the starting point that S is equal to the interior of its boundary. This problem was not as hard as I made it to be. My first time in mathstackexchange. This is amazing. – Jon E. Apr 20 '18 at 20:55