My teacher said without proving that in topological space $(X,T)$, $A\subseteq X$ is closed if and only if $b(A)\subseteq A$ where $b(A)=\bar{A}-A^{\circ}$. Is there a way to show this?
$A$ is closed if and only if it contains its boundary.
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general-topology
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0@AlexMathers, I have already looked at this answer but it doesn't answer my question. – 2017-01-28
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0Okay sorry about that, then. I've answered the question – 2017-01-28
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0Definitions are your friends, and you have been helpful in including the definition of $bd(A)$, the boundary of $A$. I think it is simple set inclusion from there. – 2017-01-28
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Well, first suppose it's closed. Then $A=\overline A$ by definition, so $b(A)=\overline{A}-A^{\circ}\subseteq \overline{A} = A$.
On the other hand, note that $\overline A=b(A)\cup A^{\circ}$. So if $b(A)\subseteq A$, then since $A^{\circ}\subseteq A$ by definition, we must have $\overline{A}\subseteq A$, i.e. $A$ is closed.