igorvanloo / dummit-foote-chapter-14-exercises Goto Github PK

Hi, I hope you're doing well. I do not see the claim of contradiction that x=σ(x) in 14.1, 1.7c, method 1. I believe the contradiction should arise from the fact that, when we choose q in the rational numbers, we have the liberty to choose that q is not be equal to σ(x), since Q is dense. Then q<x implies σ(q)<σ(x), i.e., q<σ(x). Then we have q<σ(x)<q which implies σ(x)=q, a contradiction. Let me know if I've made an oversight here, and I appreciate your contribution of these exercises! Have a good one.