discountbrains Posted September 13, 2019 Share 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 Link to comment Share on other sites More sharing options...
Strange Posted September 13, 2019 Share Posted September 13, 2019 ! Moderator Note You were told not to bring this up again. Link to comment Share on other sites More sharing options...
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