Let A ⊆ R be nonempty and bounded above, and let S be the union of all A ∈ A. (a) First, prove that S ∈ R by showing that it is a cut. (b) Now, show that S is the least upper bound for A. This finishes the proof that R is complete. Notice that we could have proved that least upper bounds exist immediately after defining the ordering on R, but saving it for last gives it the privileged place in the argument it deserves. There is, however, still one loose end to sew up. The statement of Theorem 8.6.1 mentions that our complete ordered field contains Q as a subfield. This is a slight abuse of language. What it should say is that R contains a subfield that looks and acts exactly like Q.

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