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Lemma

If $X$ is a topological space then $$ \partial(A\cap B)\subseteq[\overline{A}\cap\partial B]\cup[\partial A\cap\overline{B}] $$ for any $A,B\subseteq X$.

Corollary

If $X$ is a topological space then $$ \partial(A\setminus B)\subseteq\partial A\cup\partial B $$ for any $A,B\subseteq X$.

Proof. By the first lemma we know that $$ \partial(A\setminus B)=\partial\big(A\cap(X\setminus B)\big)\subseteq[\overline A\cap\partial(X\setminus B)]\cup[\partial A\cap\overline{X\setminus B}]=[\overline A\cap\partial B]\cup[\partial A\cap\overline{X\setminus B}]\subseteq\partial A\cup\partial B. $$ for any $A,B\subseteq X$.

So I ask if generally is $\partial (A\setminus B)=\partial A\cup\partial B$ when $B\subseteq A$ and if not I ask if with some additional hypotheses about $X$ (hausdorff separability, connectedness, etc...) or $A$ and $B$ it could be true. For example if the closure of B is contained in the interior of A then does the equality hold? So could someone help me, please?

Antonio Maria Di Mauro
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    Let $A=[0,1]^2$ and $B=[0,\frac{1}{2}]\times[0,1]$. Then... – Sangchul Lee Aug 02 '20 at 17:45
  • @SangchulLee Okay, and if $A$ is open? could be it true? – Antonio Maria Di Mauro Aug 02 '20 at 18:02
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    You may tweak the above example in such case. On the other hand, if the closure of $B$ is contained in the interior of $A$, then the equality will hold. – Sangchul Lee Aug 02 '20 at 18:06
  • @SangchulLee You said: «on the other hand, if the closure of B is contained in the interior of A, then the equality will hold». Could you prove this? I need it! – Antonio Maria Di Mauro Aug 02 '20 at 18:08
  • That follows from the formula on the last line. Then $\partial B \cap \overline{A}= \partial A$ and $\partial A \cap \overline{B}=\partial A$. – Henno Brandsma Aug 02 '20 at 18:22
  • @HennoBrandsma Umm... if $\overline B\subseteq\overset{°}A$ then $\partial B\subset\overline A$ so that surely $\partial B\cap\overline A=\partial B$ but I don't understand why $\partial B\cap\overline A=\partial A$. Then why $\partial A\cap\overline B=\partial A$? Could you explain, please? – Antonio Maria Di Mauro Aug 02 '20 at 18:25
  • Then by this why the statemet follows? – Antonio Maria Di Mauro Aug 02 '20 at 18:27
  • I meant $\partial A \cap \overline{X \setminus B} = \partial A$. – Henno Brandsma Aug 02 '20 at 18:28
  • @HennoBrandsma Excuse me, but I don't understand. Forgive my confusion. – Antonio Maria Di Mauro Aug 02 '20 at 18:32
  • If $A\subseteq B$ I know that $X\setminus B=A\setminus B\cup X\setminus A$ so that $\overline{X\setminus B}=\overline{A\setminus B}\cup\overline{X\setminus A}\subseteq\overline{A}\cup\overline{X\setminus A}$ and so $\partial A\cap \overline{X\setminus B}\subseteq\partial A\cap\Big(\overline{A}\cup\overline{X\setminus A}\Big)=\partial A$ – Antonio Maria Di Mauro Aug 02 '20 at 18:37
  • The strict inclusion in $\partial(A\setminus B)\subseteq\partial A\cup\partial B$ occurs when the two boundaries $\partial A$ and $\partial B$ overlaps, hence some cancellation takes place in $A\setminus B$. But if we assume $\overline{B}\subseteq\mathring{A}$, then $\partial B\subseteq\overline{B}$ and $\partial{A}\subseteq X\setminus\mathring{A}$ are supported on disjoint closed sets, and so, we can prevent cancellation from happening and we may expect the equality to hold. I added the actual proof below. – Sangchul Lee Aug 02 '20 at 19:01

1 Answers1

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Suppose that $B \subseteq A$. Then by using that $\overline{A_1\cup A_2} = \overline{A_1}\cup\overline{A_2}$ for any sets $A_1$ and $A_2$,

$$ \partial A\cup\partial B \subseteq \overline{X\setminus A}\cup\overline{B} = \overline{(X\setminus A)\cup B} = \overline{X\setminus(A\setminus B)}. \tag{1} $$

Now we also assume that $\overline{B}\subseteq\mathring{A}$. Then

$$ \overline{X\setminus A} \cap \overline{B} \subseteq \overline{X\setminus A} \cap \mathring{A} = \varnothing. $$

So

\begin{align*} \partial B &= \overline{X\setminus B} \cap \overline{B} \\ &= (\overline{A\setminus B} \cup \overline{X\setminus A}) \cap \overline{B} \\ &= (\overline{A\setminus B} \cap \overline{B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\ &= \overline{A\setminus B} \cap \overline{B} \end{align*}

and this shows that $\partial B \subseteq \overline{A\setminus B}$. Simiarly,

\begin{align*} \partial A &= \overline{X\setminus A} \cap \overline{A} \\ &= \overline{X\setminus A} \cap (\overline{A \setminus B} \cup \overline{B}) \\ &= (\overline{X\setminus A} \cap \overline{A \setminus B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\ &= \overline{X\setminus A} \cap \overline{A \setminus B} \end{align*}

shows that $\partial A \subseteq \overline{A\setminus B}$. Consequently

$$ \partial A \cup \partial B \subseteq \overline{A\setminus B} \tag{2} $$

and combining $\text{(1)}$ and $\text{(2)}$ proves the inclusion $\partial A \cup \partial B \subseteq \partial (A\setminus B)$.

Sangchul Lee
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  • Okay, it is clear. However if $\overline{X\setminus A}\cap\overline B=\varnothing$ then why you write $(\overline{A\setminus B}\cap\overline{B})\cup(\overline{X\setminus A}\cap\overline{B}\subseteq\overline{A\setminus B}\cap\overline{B}$ and $(\overline{X\setminus A}\cap\overline{A\setminus B})\cup(\overline{X\setminus A}\cap\overline{B})\subseteq\overline{X\setminus A}\cap\overline{A\setminus B}$? – Antonio Maria Di Mauro Aug 02 '20 at 19:31
  • Why do not write and not $(\overline{A\setminus B}\cap\overline{B})\cup(\overline{X\setminus A}\cap\overline{B}=\overline{A\setminus B}\cap\overline{B}$ and $(\overline{X\setminus A}\cap\overline{A\setminus B})\cup(\overline{X\setminus A}\cap\overline{B})=\overline{X\setminus A}\cap\overline{A\setminus B}$? – Antonio Maria Di Mauro Aug 02 '20 at 19:31
  • Anyway thanks too much for your assistance!!! – Antonio Maria Di Mauro Aug 02 '20 at 19:33
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    @AntonioMariaDiMauro, You are right. Originally I included the derivation $$\overline{X\setminus A} \cap \overline{B} \subseteq \overline{X\setminus A} \cap \mathring{A} = \varnothing$$ as part of the other two computations, and I forgot to replace $\subseteq$ by $=$ when I separated that part out. Now I changed those parts. :) – Sangchul Lee Aug 02 '20 at 19:33
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    Don't worry, it can happen! – Antonio Maria Di Mauro Aug 02 '20 at 19:35
  • Hi, could I ask your assistance here, please? – Antonio Maria Di Mauro Aug 14 '20 at 13:40