Intermediate Value Theorem and The Completeness Axiom
As we showed above, the Intermediate value theorem can be proved using the completeness axiom. In fact, the intermediate value theorem is equivalent to the completeness axiom; that is to say, any unbounded dense subset S of R to which the intermediate value theorem applies must also satisfy the completeness axiom.
To show this, take some bounded-above subset A of S. We will show that A has a least upper bound, using the intermediate value theorem. Consider the function f : S → defined by f(x) = 1 if x is an upper bound for A (i.e., for all s ∈ A, x ≥ s) and 0 otherwise. If A has only one element then that element is a least upper bound for A. Otherwise, there is some element a in A, and hence in S, which is not an upper bound for A. So f(a) = 0. Because A is bounded above and S is bounded, there exists b ∈ S such that b is an upper bound for S. So f(b) = 1. By the intermediate value theorem, if f is continuous then there must be some y ∈ S such that f(y) = ½. Since there is no such y, f is not continuous. So there is some point z in S such that for some ε > 0, for all δ > 0, f is not continuous at z.
f(z) cannot equal 0, though: if f(z) = 0 then z is not an upper bound for A, so any w < z is not an upper bound for A. So since f is discontinuous at z, there exists some point p in the range, for any δ > 0, such that p is an upper bound for A. But then this contradicts z not being an upper bound for A: suppose m ∈ A such that m > z. We know that there exists p with z < p < z + ½ ( m - z ) which is an upper bound for A (setting δ = ½ ( m - z ) ), so p > m. But p < z + ½ ( m - z ) < m, which is a contradiction.
So f(z) = 1, so z is an upper bound for A. Since for all v > z, v is an upper bound for A, f(v) = 1 for all v > z. Since f is not continuous at z, there there exists some point q in the range, for any δ > 0, such that q is not an upper bound for A. Now suppose there is some l ∈ S such that l < z and l is an upper bound for A. We then set δ to be ½( l - z ), so there is some r ∈ which is not an upper bound for A. But then l < z - ½ ( l - z ) < r, so l is not an upper bound for A, which is a contradiction.
So z is a least upper bound for A.
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