# Large Complete Ordered Fields

1. Oct 12, 2007

### Dragonfall

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

2. Oct 12, 2007

### SiddharthM

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

3. Oct 12, 2007