2

I would like to show that $$\tag{*} \partial ( A \cup B ) = \partial A \cup \partial B $$ under the condition $$\tag{**} \overline{A} \cap B = \varnothing = A \cap \overline{B} $$

This question has already been asked and correctly answered in, for instance, here.

What I'm looking for is a proof in a direct form as here, without proving the double inclusion. Is it possible? My attempts gave no positive outcome.

Something like $$ \begin{align} \partial ( A \cup B )&=\overline{A\cup B}-(A\cup B)^o\\ &= \overline{A}\cup \overline{B}-(A\cup B)^C{}^C{}^o\\ &= \overline{A}\cup \overline{B}-(A^C\cap B^C){}^C{}^o\\ &= (\overline{A}\cup \overline{B})\cap (A^C\cap B^C){}^C{}^o{}^C\\ &= (\overline{A}\cup \overline{B})\cap \overline{A^C\cap B^C}\\ \end{align} $$

or starting from the bottom $$ \begin{align} \partial A \cup \partial B &=(\overline{A}-A^o)\cup(\overline{B}-B^o)\\ &=(\overline{A}\cap A^o{}^C)\cup(\overline{B}\cap B^o{}^C)\\ &=(\overline{A}\cap \overline{A^C})\cup(\overline{B}\cap \overline{B^C}) \end{align} $$

and somewhere apply (**) to get (*).

PeptideChain
  • 829
  • If you provide your attempt it will help me understand what you mean by "a proof in a direct form". – Yanko Dec 29 '18 at 13:10
  • @Yanko I would like a sequence of equalities; I edited the question – PeptideChain Dec 29 '18 at 13:27
  • A proof is a proof. Why does it have to be a chain of equalities? Maybe you're too stuck on set algebra? – Henno Brandsma Dec 29 '18 at 17:32
  • @HennoBrandsma In my option would be a nicer proof. I can do also without it. However the condition (**) is so simple that I imagined that it could be replaced in some expansion of one of the two sides of the equation. If someone find the proof is good, otherwise it is also good. – PeptideChain Dec 30 '18 at 10:41

0 Answers0