# Every ordered field that has the least upper bound property is isomorphic to the real number system.

Okay, so here's a theorem from Rudin:

"Every ordered field that has the least upper bound property is isomorphic to the real number system."

Here's a definition:

"Ordered fields are isomorphic if there is a bijection between the underlying sets that preserves the field operations and the orders."

i.e.

Ordered fields (F, +, ·, ≤) and (K, ⊕, ⊙, ≼) are isomorphic if there exists a bijection h : F → K such that (a) for all x,y ∈ F, h(x+y)=h(x)⊕h(y), (b) for all x,y ∈ F, h(x·y)=h(x)⊙h(y), and (c) for all x,y ∈ F, if x≤y,then h(x)≼h(y).

Here are some things that I think are true:

1.) The field T with two elements {0, 1} is an ordered field that has the least upper bound property.

Because T is finite, every nonempty subset of T has a maximum. Max(T) = LUB(T) for all subsets of T that have a maximum element. Therefore all subsets of T have a least upper bound in T. Thus T has the least upper bound property.

2.) Since T has the LUB property, it is isomorphic to the real # system and therefore there exists a bijection between the real numbers and {0, 1}.

3.) ...but since {0, 1} is finite and the real numbers are infinite, there can be no surjection from {0,1} -> real numbers. Therefore there's no bijection between the reals and {0, 1}.

So... Which thing that I think is true is not actually true?

Thanks!

• The field with two elements is not an ordered field; $1 + 1$ is $0$ in that field, but it would have to be greater than $0$ in an ordered field. Every ordered field has characteristic $0$, actually. See en.wikipedia.org/wiki/Ordered_field – Carl Mummert Aug 30 '15 at 2:48
• Ah! Great! Thanks! – boxTurtle Aug 30 '15 at 2:51
• Ah hurray! And my next question was going to be "What about the integers then?" But that's not an ordered field either, because that's not a field! – boxTurtle Aug 30 '15 at 2:57