Definition D1
c is the supremum of S if c is an upper bound of S and no number smaller than c is an upper bound of S.
Proof.
(1)An upper bound is a number not smaller than any element of S. The supremum is the smallest upper bound.convention
∎
Definition D2
f is continuous at p if for every ε > 0 there is δ > 0 such that x ∈ [a,b] and |x − p| < δ imply |f(x) − f(p)| < ε.
Proof.
(1)δ may depend on p and ε. This definition is a property at the single point p.convention
∎
Lemma L1
Let S = {x ∈ [a,b] : f(x) < k}. Then c = sup S exists and a ≤ c ≤ b.
Proof.
(1)f is continuous on [a,b] and f(a) < k < f(b).assumption
(2)Since f(a) < k, we have a ∈ S, so S ≠ ∅.(1), convention
(3)Since S ⊆ [a,b], b is an upper bound of S.convention, D1
(4)Therefore c = sup S exists.(2), (3), A7
(5)Since a ∈ S and c is an upper bound, a ≤ c.(2), (4), D1
(6)Since b is an upper bound and c is the least upper bound, c ≤ b.(3), (4), D1
∎
Lemma L2
f(c) < k does not hold.
Proof.
(1)Suppose f(c) < k.assumption
(2)Since f(b) > k, c ≠ b, and since c ≤ b, c < b.(1), L1, A3
(3)Set ε = k − f(c). Then ε > 0.(1), A1, A2
(4)Since c ∈ [a,b], there is δ > 0 such that |x − c| < δ implies |f(x) − f(c)| < ε.L1, D2, assumption
(5)b − c > 0.(2), A1, A2
(6)Choose h with 0 < h < δ, h < b − c.(4), (5), A6
(7)Set x₁ = c + h. Then c < x₁ < b and x₁ ∈ [a,b].(6), L1, A1, A2, A4
(8)x₁ − c = h and −δ < h < δ, so |x₁ − c| < δ.(6), (7), A5
(9)|f(x₁) − f(c)| < ε, so f(x₁) < f(c) + ε = k.(4), (8), A5, (3), A1
(10)Therefore x₁ ∈ S.(7), (9), convention
(11)Since x₁ > c, c is not an upper bound of S. This contradicts D1.(7), (10), D1, L1
(12)f(c) < k does not hold.(11)
∎
Lemma L3
f(c) > k does not hold.
Proof.
(1)Suppose f(c) > k.assumption
(2)Since f(a) < k, c ≠ a, and since a ≤ c, a < c.(1), L1, A3
(3)Set ε = f(c) − k. Then ε > 0.(1), A1, A2
(4)Since c ∈ [a,b], there is δ > 0 such that |x − c| < δ implies |f(x) − f(c)| < ε.L1, D2, assumption
(5)c − a > 0.(2), A1, A2
(6)Choose h with 0 < h < δ, h < c − a, and set c′ = c − h. Then a < c′ < c.(4), (5), A6, A1, A2
(7)Suppose x ∈ S and x > c′.assumption
(8)Since c is an upper bound, x ≤ c, and since x > c − h, we have −δ < x − c ≤ 0 < δ.(6), (7), L1, D1, A1, A2
(9)Since |x − c| < δ, we have |f(x) − f(c)| < ε.(8), A5, (4)
(10)f(x) > f(c) − ε = k.(9), A5, (3), A1
(11)Since x ∈ S, f(x) < k. This contradicts (10).(7), convention, (10), A3
(12)Therefore every element of S is at most c′. Then c′ is an upper bound smaller than c, contradicting D1.(11), (6), D1, L1
(13)f(c) > k does not hold.(12)
∎
Theorem T
If f is continuous on [a,b] and f(a) < k < f(b), then there exists c between a and b with f(c) = k.
Proof.
(1)c = sup S exists and a ≤ c ≤ b.L1
(2)Neither f(c) < k nor f(c) > k holds.L2, L3
(3)Therefore f(c) = k.(2), A3
(4)Since f(a) < k = f(c), c ≠ a, and since f(b) > k = f(c), c ≠ b.assumption, (3)
(5)Since a ≤ c ≤ b, we have a < c < b.(1), (4), A3
∎