Proof
The theorem may be proved as a consequence of the completeness property of the real numbers as follows:
We shall prove the first case f(a) < u < f(b); the second is similar.
Let S be the set of all x in such that f(x) ≤ u. Then S is non-empty since a is an element of S, and S is bounded above by b. Hence, by completeness, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f(c) = u.
- Suppose first that f(c) > u, then f(c) − u > 0. Since f is continuous, there is a δ > 0 such that | f(x) − f(c) | < ε whenever | x − c | < δ. Pick ε = f(c) − u, then | f(x) − f(c) | < f(c) − u. But then, f(x) > f(c) − (f(c) − u) = u whenever | x − c | < δ (that is, f(x) > u for x in (c − δ, c + δ)). This requires that c − δ be an upper bound for S (since no point in the interval (c − δ, c] for which f > u, can be contained in S, and c was defined as the least upper bound for S), an upper bound less than c. The contradiction nullifies this paragraph's opening assumption.
- Suppose instead that f(c) < u. Again, by continuity, there is a δ > 0 such that | f(x) − f(c) | < u − f(c) whenever | x − c | < δ. Then f(x) < f(c) + (u − f(c)) = u for x in (c − δ, c + δ). Since x=c + δ/2 is contained in (c − δ, c + δ), it also satisfies f(x) < u, so it must be contained in S. However, it also exceeds the least upper bound c of S. The contradiction nullifies this paragraph's opening assumption, as well.
We deduce that f(c) = u as stated.
An alternative proof may be found at non-standard calculus.
Read more about this topic: Intermediate Value Theorem
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