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?

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Anonymous Sun Jun 11 20:18:28 2017 No.8970823 [Reply] [Original]

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?

>>	Anonymous Sun Jun 11 20:24:01 2017 No.8970832 >>8970823 No and no Show me an inductive construction of the set of real numbers

Anonymous Sun Jun 11 20:43:31 2017 No.8970853

>>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.

>>	Anonymous Mon Jun 12 03:18:00 2017 No.8971256 >>8970832 >disproving well-ordering theorem with just one word

>>	Anonymous Mon Jun 12 03:21:19 2017 No.8971260 >>8970853 Vitali set is constructed using transfinite incuction and well-ordering of reals, so it's clearly not impossible

>>	Anonymous Mon Jun 12 07:15:14 2017 No.8971507 >>8970823 (0,1) is a subset of R with no least element. R is not well-ordered.

>>	Anonymous Mon Jun 12 07:18:17 2017 No.8971512 >>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

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