User:Alec/Things not to forget/Q9 wreckage

See:

• Exercises:Mond - Topology - 1/Question 9

Problem body

• • • • • • • • • So we have shown: $(\neg(\text{Disjoint}))\implies(\forall p\in \pi(U_a)\cap\pi(U_b)\exists q\in \pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))[\pi(q)=p])$ and by tidying up: $(\neg\text{Disjoint})\implies(\forall p\in \pi(U_a)\cap\pi(U_b)\exists q\in U_a\cap U_b[\pi(q)=p])$
                • By contrapositive:
                  • $\big[(\neg\text{Disjoint})\implies(\forall p\in \pi(U_a)\cap\pi(U_b)\exists q\in U_a\cap U_b[\pi(q)=p])\big]\iff\big[\neg(\forall p\in \pi(U_a)\cap\pi(U_b)\exists q\in U_a\cap U_b[\pi(q)=p])\implies\neg(\neg\text{Disjoint})\big]$
                • Arriving at: $\neg(\forall p\in \pi(U_a)\cap\pi(U_b)\exists q\in U_a\cap U_b[\pi(q)=p])\implies \text{Disjoint}$

Fix failed:

• • • • • • • • clearly $\exists q\in \pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))$ such that $\pi(q)=p$
                • However $\pi^{-1}(\pi(U_a))\cap\pi^{-1}(\pi(U_b))=U_a\cap U_b$ and $U_a\cap U_b=\emptyset$ (by construction), so there does not exist such a $q$!
                • If there is no $q\in U_a\cap U_b$ such that $\pi(q)=p$ then $p\notin\pi(U_a)\cap\pi(U_b)$