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The question is $:$

Let $(X,d)$ be a metric space.Let $A,B \subseteq X$.Then prove that $\partial (A \cup B)= \partial(A) \cup \partial(B)$ if $\bar {A} \cap \bar {B} = \phi$ , where $\partial (A)$ and $\bar A$ respectively denote the boundary of $A$ and closure of $A$ in $(X,d)$.

My attempt $:$

Let $A^{o}$ denote the interior of $A$.Then we know that $\partial(A) = \bar A \setminus A^{o}$. Now,$\partial (A \cup B) = \overline {A \cup B} \setminus (A \cup B)^{o} = (\bar A \cup \bar B) \setminus (A \cup B)^{o} = (\bar A \setminus (A \cup B)^{o}) \cup (\bar B \setminus (A \cup B)^{o}) \subseteq (\bar A \setminus A^{o}) \cup (\bar B \setminus B^{o}) = \partial(A) \cup \partial(B)$.

But I fail to prove the other part of the inequality using the given condition. Please help me in proving it.

Thank you in advance.

Arnab Chattopadhyay.
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2 Answers2

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Suppose $x\in \partial (A)\cup\partial(B)$. Assume WLOG $x\in \partial (A)$ (because the two boundaries are disjoint). Then $\forall \epsilon>0, B(x,\epsilon)\cap A\neq \emptyset\ \text{and}\ B(x,\epsilon)\cap(X\setminus A)\neq\emptyset$. Therefore $\forall \epsilon>0, B(x,\epsilon)\cap (A\cup B)\neq\emptyset$. Assume $\exists\epsilon_0>0$ such that $B(x,\epsilon_0)\cap (X\setminus(A\cup B))=\emptyset$. Then $B(x,\epsilon_0)\subseteq A\cup B$. Therefore $x\in\ \text{int} (A\cup B)$. Therefore $x \in\text{int}B$. So $\exists \epsilon_1>0$ such that $B(x,\epsilon_1)\subseteq B$. Therefore $B(x,\epsilon_1)\cap A =\emptyset$; contradiction. So we have $\forall\epsilon>0,B(x,\epsilon_0)\cap (X\setminus(A\cup B))\neq\emptyset$. Hence $x\in\partial(A\cup B)$.

Janitha357
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  • can you make it clear that why does $x \in int (A \cup B)$ imply $x \in int (B)$? – Arnab Chattopadhyay. May 14 '17 at 11:33
  • $x$ is a boundary point of $A$. – Janitha357 May 14 '17 at 11:39
  • Then what will happen?Do you assume that $int (A \cup B) = int (A) \cup int (B)$? – Arnab Chattopadhyay. May 14 '17 at 12:02
  • According to my logic since $B(x,\epsilon_{0}) \subseteq A \cup B$.Take any $r \leq \epsilon_{0}$ then clearly $B(x,r) \subseteq A \cup B$.But $B(x,r)$ is not entirely contained in $A$ for otherwise $x \in int (A)$ which contradicts the fact that $x \in \partial A$.Hence for any $r \leq \epsilon_{0}$ and hence for any $r >0$ , $B(x,r) \cap B \neq \phi$.$\implies$ $x \in \bar B$.So $x \in \bar {A} \cap \bar {B}$ , a contradiction. – Arnab Chattopadhyay. May 14 '17 at 12:44
  • @ArnabChatterjee You are correct in your doubt. $x \in $ interior of $A \cup B $ does not alone in general imply that $x \in $ interior of $B $. For consider $A = [0,2] $,$B = [1,3] $ under the usual metric. Then $x=1$ is in the interior of $A \cup B $, but not in the interior of $B $. – bloomers May 24 '17 at 09:42
  • @bloomers your comment is just exactly what I am looking for.Thanks for your help.Is my previous logic correct?Please verify it. – Arnab Chattopadhyay. May 25 '17 at 11:10
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@ArnabChatterjee Your logic looks sound. Excellent reasoning!

A ${slightly}$ more direct approach you might consider could be as follows:

Let $ x \in \partial{A}\cup\partial{B}.$ WLOG $ x \in \partial{A}$ (just as before). Since $cl(A) $ and $cl(B) $ are disjoint, then $cl(A) \subseteq X \backslash cl(B)$, which is open. Then there must be an $\epsilon_0$ that keeps $B(x,\epsilon_0) $ disjoint from $cl(B)$. Since $x$ is in the boundary of $A$, then $B(x,\epsilon_0)$ has non-empty intersection with $ A \subseteq A \cup B$, as well as non-empty intersection with $X \backslash A$. It follows from such $\epsilon_0$ that any $y$ in $B(x,\epsilon_0)\backslash A$ cannot be in $B$, so such $y$ is also in $X\backslash B $; that is to say that $B(x,\epsilon_0) \cap \color{red}{(X\backslash A)\cap (X\backslash B)}$$ \:(which\:is\: \color{red}{X\backslash(A\cup B)})$ is also non-empty. Hence, for any $\epsilon \leq \epsilon_0 $ (i.e. $\forall \epsilon > 0$), we have $B(x,\epsilon)\cap(A\cup B)\neq \emptyset$, and $B(x,\epsilon)\cap [X\backslash(A\cup B)]\neq \emptyset$.

bloomers
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