On Sat, 22 Jun 2013 09:56:15 -0700 (PDT), firstname.lastname@example.org wrote:
>On Saturday, 22 June 2013 17:44:42 UTC+2, dull...@sprynet.com wrote: >> Consider the following statement: (*) "In order to order the reals, you must be able to distinguish them. But it is impossible to distinguish more than countably many." _Is_ (*) correct? > >Yes, that is correct. > >> If (*) is correct it follows that there's no _order_ on the reals, which seems curious, to say the least, since there _is_ a perfectly simple, standard order on the >>reals, nothing to do with AC or anything else problematic. > >There is an order on the rationals. In order to put an irrational into that order, usually its rational approximations are chosen. By the way, this is another reason >that not more irrationals than rationals can exist. Every pair of irrational numbers must be distinguished by a rational between them.
Oh! You say that every pair of irrationals "must" be distinguished by a rational between them. You say we need to do this in order to define the standard order on the reals. And you agree that there is no problem defining the standard order on the reals.
So you must agree that any two reals _can_ be distingushed by a rational between them.
In particular, and two reals _can_ be distingushed.
Now what the heck are you talking about when you say there's no well-order on the reals because we cannot distinguish them?
> >I think you believe that we have the complete set of real numbers when we draw an axis of the cartesian coordinate system? >No, then we only distribute some atoms of carbon or molecules of chalk. Nothing else. An irrational number is given by a finite formula >like SUM(1/n!). That formula enables us associate the irrational number with an arbitrarily small interval on the axis. > >Regards, WM