Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ???

share|improve this question
    
Be carefull, compactness iff it is closed and bounded is not true in arbitrary metric spaces. –  Gastón Burrull Feb 24 '13 at 18:12

1 Answer 1

up vote 3 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.