# Closure of boundary a boundary of closure.

Were ment to show that $$$$\overline {\delta A}=\delta \bar{A}$$$$ $$\bar{A}$$ being the closure, $$A^\circ$$ being the interior and $$\delta$$A being the boundary. I've tried doing it at such: \begin{align} \overline {\delta A}=\delta A \cup\delta A=\delta A\\ \end{align} \begin{align} \delta\bar{A}=\delta(A\cup\delta A)=\overline {A\cup\delta A}\backslash(AU\delta A)^\circ=A\cup\delta A \backslash A^\circ \end{align} Now the last equal sign must be larger than $$\delta A$$ since $$\delta A$$ \ $$A^\circ$$ is an empty set, I am adding something to $$\delta A$$, hence it must be larger than $$\delta A$$. Yet it seems wrong for example if I Drive a set in R$$^2$$ with a missing line, the inclusion doesn't seem to work. Any help appreciated.

This is false. If $$A=\mathbb Q$$ then $$\delta (A)=\mathbb R$$ and $$\delta \overline {A} =\emptyset$$,
• What is the closure of $\mathbb R$ and what is its interior? @Mark Commented May 8, 2020 at 11:54