Are There Complete Ordered Fields Larger Than 2^{\aleph_0}?

[Dragonfall]

I was told that there are no complete ordered fields of cardinality greater than $$2^{\aleph_0}$$. Why is that?

 

[SiddharthM]

i'm not 100% sure about this but..

I remember hearing a theorem that says that every ordered field with the least upper bound property is isomorphic to the reals. So every ordered field that is order-complete (ie has least upper bound property) has to have cardinality c.

hmmm..i'm not sure order-complete and complete are always equivalent

 

[matt grime]

Let's think about it:

A field has to have 0 and 1, and hence all things of the form a/b for a,b sums of 1. This has cardinality at most |Q| the rationals, i.e. aleph_0. Now it has to be complete, so it must contain limits of all sequences of these elements. There are 2^\aleph_0 of these. The only question is now if there can be any other element. No there can't - if there were some element x I've not described, then I can assume it is positive by looking at -x if necessary, and if it's larger than 1, I can replace it with 1/x. Thus I have to show that any x such that 0<x<1 has already been described. But this is true, since I can construct a sequence of 'rationals' that approximate it by repeated bisection of the interval.