Prove that in any metric space $(X,\rho)$ for any $S\subset{X}$, we have $bd(bd(S)) = bd(bd(bd(S)))$, while not necessarily $bd(S)=bd(bd(S))$

Prof's Hint (first show that a boundary of a closed set has an empty interior)

By Heine-Borel, I know a subset of $\mathbb{R}^n$ is compact iff it is closed and bounded.

Also a subset $S$ of $\mathbb{R}^n$ is called bounded if there is a real number $R$ such that $S$ is contained in the ball $B_R (0)$

So, $bd(S)$ must give us the information the $S$ is empty.

Intuitively, boundary of the boundary equals to the boundary. (Its like there is no place we can go any further)

But the question clear stated it is not necessary ???

  • $\begingroup$ Be carefull, compactness iff it is closed and bounded is not true in arbitrary metric spaces. $\endgroup$ – Gaston Burrull Feb 24 '13 at 18:12

Heine-Borel has nothing to do with it. Just use basic definitions of closure, boundary, interior.

Some key facts: if a set $B$ is closed and has empty interior, $\partial B = B$. Proof: suppose $x \in B$. Then for all $r > 0$, $B(x,r)$ is not a subset of $B$ (otherwise $x$ would be an interior point of $B$ and the interior is empty). This means that every ball $B(x,r)$ around $x$ intersects $X \setminus B$ and also it intersects $B$ (in $x$), so $x \in \partial B$. For the other inclusion: suppose $x \in \partial B$. Then every ball $B(x,r)$ around $x$ intersects $B$ (as well as $X \setminus B$) so $x$ is in the closure of $B$ which equals $B$ as $B$ is closed. So $x \in B$ and we have equality.

Your professor's hint was thus to show that $\partial \partial A$ is such a closed set with empty interior, so it equals its boundary.

Fact 2: the boundary $\partial B$ of any set $B$ is closed. This is clear from the fact that $\partial B = \overline{B} \cap \overline{X \setminus B}$, which is the intersection of closed sets and thus closed. The equality as stated is just a refomulation of the definition (or itself the definition, depending on the text) that a point is a boundary point of $B$ iff every open ball around it intersects both the set and its complement.

Fact 3: The boundary of a closed set $C$ has empty interior. First note that $\partial C \subset C$ for a closed set, as is clear from the equality from fact 2: $\partial C = \overline{C} \cap \overline{X \setminus C} \subset \overline{C} = C$. So suppose the interior of $\partial C$ is not empty, then there is a point $x \in \partial C$ and an $r>0$ such that $B(x,r) \subset \partial C (\subset C)$. But as $x$ is in $\partial C$ in particular every ball around $x$ should intersect $X \setminus C$ as well, which cannot be as it lies entirely inside $C$, so we have a contradiction.

So $\partial A$ is closed (fact 2) so $\partial \partial A$ is closed too and moreover has empty interior by fact 3. So we apply fact 1 to see that $\partial \partial \partial A = \partial \partial A$, as required.


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