User:Alec/Things not to forget/Q9 wreckage

See:

• Exercises:Mond - Topology - 1/Question 9

Problem body

• • • • • • • • • So we have shown: [ilmath](\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])[/ilmath] and by tidying up: [ilmath](\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])[/ilmath]
                • By contrapositive:
                  • [ilmath]\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][/ilmath]
                • Arriving at: [ilmath]\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}[/ilmath]

Fix failed:

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