Let $(X,\rho)$ be a metric space and $A\subset X$. Prove that $\partial A = \overline A \cap \overline{X \setminus A} $. I have no idea how to prove that. Please help.

  • $\begingroup$ How is teh boundary defined? As the closure less the interior? $\endgroup$ – ncmathsadist May 28 '13 at 23:58
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    $\begingroup$ Not every question including the word "set" is a question in set theory. $\endgroup$ – Asaf Karagila May 28 '13 at 23:58
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    $\begingroup$ What definition of $\partial A$ are you using? Have you solved other problems where you have to show that two sets are equal? $\endgroup$ – Jonas Meyer May 28 '13 at 23:59
  • $\begingroup$ C'est vrai, Asaf. $\endgroup$ – ncmathsadist May 28 '13 at 23:59
  • $\begingroup$ My definition is that $\partial A$ is the set of all points all of whose neighborhoods meet both $A$ and $A^c$. $\endgroup$ – ncmathsadist May 29 '13 at 0:00

To show two sets are equal, first assume we have an element of the first set, and show it is in the second set. Then, assume we have an element of the second set, and show it is in the first.

Proof outline:

  1. Let $x \in \partial A$ arbitrary. Show that $x \in \overline{A}$ (this should be easy from your definition). Then, show that $x \in \overline{X \setminus A}$ using the fact that $\partial A = \partial (X \setminus A)$.
  2. Let $y \in \overline{A} \cap \overline{X \setminus A}$ arbitrary. Show that $y \in \partial A$ by showing that $y \in \overline{A}$ but $y \not \in \text{Int } A$.
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    $\begingroup$ "using the fact that $\partial A=\partial(X\setminus A)$": Also proving that fact, of course:) $\endgroup$ – Jonas Meyer May 29 '13 at 0:40
  • $\begingroup$ @JonasMeyer certainly :) See also my comment on the question itself. $\endgroup$ – 6005 May 29 '13 at 0:45

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