Definition D1
f is continuous at a point p of a set I means that for every ε > 0 there is δ > 0 such that if x ∈ I and |x − p| < δ, then |f(x) − f(p)| < ε.
Proof.
(1)ε and δ are taken from the numbers of the stage, that is, from the rationals. The stage contains no other numbers.convention
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Definition D2
Let I = {x ∈ ℚ : 1 ≤ x ≤ 2}, and on I let f(x) = x² − 2.
Proof.
(1)If x ∈ ℚ, then x² − 2 ∈ ℚ, so all values of f lie on the stage.A1
(2)This stage has no completeness axiom. All that may be used are the field axioms and the order axioms.convention
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Lemma L1
If x, p ∈ I, then |x + p| ≤ 4.
Proof.
(1)x ∈ I, so 1 ≤ x ≤ 2.D2
(2)0 < 1, so −2 < 0 < 1 ≤ x, and therefore −2 ≤ x ≤ 2, that is, |x| ≤ 2.(1), A7, A2, A5, A8
(3)p ∈ I, so |p| ≤ 2 for the same reason.(2), D2
(4)|x + p| ≤ |x| + |p| ≤ 2 + 2 = 4(2), (3), A11, A2, A5, A1
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Lemma L2
f is continuous at every point of I.
Proof.
(1)Suppose p ∈ I and ε > 0 are given.assumption, D2
(2)Let δ = ε/4. ℚ is a field, so δ ∈ ℚ, and 4 > 0, so δ > 0.(1), A1, A7, A2, A6, A3
(3)Suppose x ∈ I and |x − p| < δ.assumption
(4)f(x) − f(p) = (x² − 2) − (p² − 2) = x² − p² = (x + p)(x − p)D2, A1, E
(5)|f(x) − f(p)| = |x + p| · |x − p|(4), A10, E
(6)|x − p| ≥ 0 and |x + p| ≤ 4, so |f(x) − f(p)| ≤ 4|x − p| < 4δ = ε(5), (3), (2), L1, A9, A4, A3, A1
(7)For every ε > 0 such a δ exists, so f is continuous at p.(6), D1
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Lemma L3
f(1) < 0 < f(2) and 1, 2 ∈ I.
Proof.
(1)f(1) = 1 × 1 − 2 = −1.D2, A1
(2)Adding −1 to both sides of 0 < 1 gives −1 < 0.A7, A2, A1
(3)Adding 1 to both sides of 0 < 1 gives 1 < 2, and by transitivity 0 < 2.A7, A2, A5, A1
(4)f(2) = 2 × 2 − 2 = 2.D2, A1
(5)Therefore f(1) = −1 < 0 < 2 = f(2).(1), (2), (3), (4)
(6)1 ≤ 1 ≤ 2 and 1 ≤ 2 ≤ 2, so 1 and 2 are elements of I.(3), D2
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Theorem T
f is continuous on I and f(1) < 0 < f(2), yet there exists no c ∈ I with f(c) = 0.
Proof.
(1)f is continuous at every point of I.L2
(2)1, 2 ∈ I and f(1) < 0 < f(2).L3
(3)Suppose c ∈ I and f(c) = 0.assumption
(4)c² − 2 = 0, so adding 2 to both sides gives c² = 2.(3), D2, A1, E
(5)c ∈ I ⊆ ℚ, so this contradicts A12.(4), (3), D2, A12
(6)Therefore there is no c ∈ I with f(c) = 0.(5)
(7)(1)(2)(6) hold together, so the hypotheses of the intermediate value theorem are satisfied and its conclusion is false.(1), (2), (6)
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Corollary C
The intermediate value theorem cannot be proved from the field axioms and the order axioms alone.
Proof.
(1)Suppose the intermediate value theorem is proved from the field axioms and the order axioms alone.assumption
(2)ℚ satisfies all of those axioms, so that proof holds unchanged in ℚ.(1), A1, A2, A3, A4, A5, A7
(3)But the f of T satisfies the hypotheses and fails the conclusion.T
(4)Contradiction. Therefore the intermediate value theorem needs one more thing that is not among these axioms.(2), (3)
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