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?
Best Answer
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 $\partial A \cup \partial B \subseteq \partial (A\setminus B)$.