discountbrains 33 Posted September 13, 2019 My first of several proofs: Consider any total linear ordering, <*, of the reals. To make it simpler consider <* for S={x: 0<x}. At this point we don't know if <* is a well ordering or not. I will show by math induction that a well ordering of S must produce a countable number of minimums for a particular collection of subsets of S. Then I'll show all numbers, z, must be in this collection or set of minimums. Thus, the conclusion must be that if R can be well ordered it must be a countable set and we know this is not true. The above is a preliminary test before going further to make sure my topic does not get closed 0 Share this post Link to post Share on other sites

Strange 4192 Posted September 13, 2019 ! Moderator Note You were told not to bring this up again. 0 Share this post Link to post Share on other sites