[FOM] Characterization of R/Simple proof

[Harvey Friedman]

As an FOM Editor, I consistently oppose the posting of proofs. I oppose this
FOM posting (smile). In light of the Enayat posting of 2/14/05 8:44PM, I
will not press my opposition to this FOM posting (smile).

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This seems to be particularly simple. No use of ordinals or transfinite
induction or transfinite recursion is used.

THEOREM 1. Let X be a linearly ordered set with left and right endpoints and
the least upper bound property. Let f:X2 into X be continuous, where for all
x < y, x < f(x,y) < y. Then X is isomorphic to the usual closed unit
interval. 

Proof: Let A be the least set containing the left endpoints and closed under
f. Obviously A is countable. Assume A is not dense. Let (x,y) be a maximal
nonempty open interval disjoint from A. We now define a1 <= a2 <= ... < x
from A, and b1 >= b2 >= ... > y from A as follows. Suppose
a1,...,ak,b1,...,bk have been defined. If f(ak,bk) < x then set ak+1 =
f(ak,bk), bk+1 = bk. If f(ak,bk) > y then set ak+1 = ak, bk+1 = f(ak,bk).
Let a* be the sup of the a's, and b* be the inf of the b's. Now infinitely
many f(ak,bk) = ak+1 or infinitely many f(ak,bk) = bk+1. In the former case,
by continuity, f(a*,b*) = a*. In the latter case, by continuity, f(a*,b*) =
b*. Both are impossible. Hence A is dense, and so X is separable. QED

THEOREM 2. Let X be a linearly ordered set with no endpoints and the least
upper bound property. Let f:X2 into X be continuous, where for all x < y, x
< f(x,y) < y. Then X is isomorphic to the reals.

Proof: By Theorem 1, every interval in X with endpoints is isomorphic to the
usual closed unit interval. Hence it suffices to prove that there is an
unbounded sequence of points in X. We define a1 < a2 < ... as follows. Let
a1 be arbitrary. Suppose a1,...,ak have been defined, k >= 1. Let ak+1 be
the sup of all f(ak,x), x > ak, if this sup exists.

Suppose ak+1 does not exist. Then the f(ak,x), x > ak, are unbounded. Define
ak < b1 < b2 ... inductively such that each f(ak,bi+1) > bi. We claim that
the b's are unbounded. Suppose not, and let b* be the sup of the b's. By
continuity, f(ak,b*) is at least as large as every bi, and so f(ak,b*) >=
b*, which is a contradiction. Thus if ak+1 does not exist then we obtain an
unbounded sequence as required. Thus we may assume without loss of
generality that each a1,a2,... exists.

Finally, we claim that the a's are unbounded. Suppose not, and let a* be the
sup of the a's. Let a* < b. Then f(a*,b) > a*. By continuity, some f(ak,b) >
a*. But f(ak,b) <= ak+1 < a*. QED

Harvey Friedman