general topology on boundary

Question

If $A$ and $B$ are open in $X$. To show that $(A\cap \delta(B))\cup (B\cap \delta(A))\subset \delta (A\cap B)\subset (A\cap \delta(B))\cup (B\cap \delta(A)) \cup (\delta(A)\cap \delta(B))$. Also state an example on real line such that these 3 sets are different. It is easy see the picture on a plane $R^2$, but on $R$ how will it look? $\delta$- denotes the boundary of set.

Answer 1

$\def\ol{\overline}$ Since for each set $Y$, $\delta(Y)=\ol{Y}\setminus\operatorname{int} Y$, we have $\delta(Y)=\ol{Y}\setminus Y$ for each open set $Y$.

Thus

$$(A\cap \delta(B))\cup (B\cap \delta(A)) \cup (\delta(A)\cap \delta(B))=$$ $$(A\cap (\ol{B}\setminus B))\cup (B\cap (\ol{A}\setminus A)) \cup ((\ol{A}\setminus A)\cap (\ol{B}\setminus B))=$$ $$\ol{A}\cap\ol{B}\setminus (A\cap B)\supset \ol{A\cap B}\setminus (A\cap B)=\delta(A\cap B).$$

Let now $x\in A\cap \delta(B)$ be any point and $U\subset A$ be any neighborhood of the point $x$. Since $x\in \delta(B)\subset \ol B$, there exist a point $y\in U\cap B\subset A\cap B$. Thus $x\in\ol{A\cap B}$. On the other hand, $$x\in\delta(B)\subset X\setminus B\subset X\setminus (A\cap B).$$ So $$x\in \ol{A\cap B}\setminus (A\cap B)=\delta(A\cap B).$$ Similarly we show that $B\cap \delta(A)\subset \delta(A\cap B)$.

Now let $X$ be the real line endowed with the usual topology, $A=(-1,0)\cup (2,3)$, and $B=(0,1)\cup (2,3)$. Then $\delta(A)=\{-1,0,2,3\}$, $\delta(B)=\{0,1,2,3\}$, $A\cap B=(2,3)$. Thus

$$(A\cap \delta(B))\cup (B\cap \delta(A))=\varnothing\ne \{2,3\}=\delta(A\cap B),$$ $$\delta(A\cap B)=\{2,3\}\ne \{0,2,3\}=(A\cap \delta(B))\cup (B\cap \delta(A)) \cup (\delta(A)\cap \delta(B)).$$