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Let $(X,T)$ be a topological space and let $S$ be a subset of $X$. Prove $Bd(S) = Cl(Bd(S))$.

My initial thought is that the boundary is a closed set of points and the closure of a closed set is going to equal itself.

Am I approaching this problem correctly?

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Yes you are right, the boundary of $S$ is defined as

$$Bd(S)=\overline{S}\setminus Int(S)=\overline{S}\cap (X\setminus Int(S))$$ which is the intersection of two closed sets and thus closed.

Therefore, $Cl(Bd(S))=Bd(S)$ like what you claimed.

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This can also be proved directly from the definitions of closure and boundary. A point $Q$ is in $bd(S)$ if every open set containing $Q$ contains some point $x\in S$ and some point $y$ not in $S$. A point $P$ is in $cl(bd(S))$ provided every open set containing $P$ contains a point of $bd(S)$. Every set in $bd(S)$ has this property, so $ bd(S)\subset cl( bd(S))$. So one must then show that $cl(bd(S))\subset bd(S)$. But if $P\in cl(bd(S))$ then it must contain some point $Q$ in $bd(S)$ and therefore also some point $x\in S$ and some point $y$ not in $S$. Therefore, $P$ must be in $bd(S)$. So $cl(bd(s))=bd(S)$.

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Yes you are right. You could also use the fact that $Bd(Bd(A)) \subseteq Bd(A)$ holds for all $A\in (X,T)$ as I do below:

To prove the equality we show that both sides are subsets of each other.

$Bd(S)\subseteq Cl(Bd(S))$ clearly holds.

The other direction follows from $Cl(Bd(S)) = Bd(Bd(S)) \cup Bd(S) \subseteq Bd(S) \cup Bd(S) = Bd(S)$

Thus the equality holds.

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  • $\begingroup$ How would you prove that Bd(Bd(A)) is a subset of Bd(A)? $\endgroup$ – NUG May 3 '16 at 10:29
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    $\begingroup$ Let $x\in bd(bd(A))$. Suppose $x\notin bd(A)$, then $x\in bd(A)^C =ext(A)\cup int(A)$, which is open. Thus there is a nbhd $U$ containing $x$ such that $U\subset ext(A)\cup int(A)$. Since $x\in bd(bd(A))$, by the definition of boundary, every nbhd of $x$ contains at least one point in $bd(A)$. Thus $U$ contains a point in $bd(A)$. Contradiction. $\endgroup$ – M47145 May 3 '16 at 17:40

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