Every set can be well-ordered, and theorems about well-ordered sets can be proved by the means of induction. So can we define well-order on [math]\left\{z\in\textbf{C}: 0<\Re(z)<\frac{1}{2}\right\}[/math] and prove Riemann hypothesis inductively?
>>8970823
No and no
Show me an inductive construction of the set of real numbers
>>8970832
https://en.wikipedia.org/wiki/Transfinite_induction
OPs description of a proof works at a very high abstract level. But the details are surely impossible.
A well ordering of that set isn't going to be anything friendly for an inductive proof.
>>8970832
>disproving well-ordering theorem with just one word
>>8970853
Vitali set is constructed using transfinite incuction and well-ordering of reals, so it's clearly not impossible
>>8970823
(0,1) is a subset of R with no least element.
R is not well-ordered.
>>8971507
>For every set X, there exists a well-ordering with domain X.
https://en.wikipedia.org/wiki/Well-ordering_theorem
(R, [math]\leq[/math]) is not well-ordered, but there exist well-ordering of R