The content in today's blog is taken from Edwards & Penney's Calculus and Analytic Geometry.
Theorem: Intermediate Value Property of Continuous Functions
If a function f is continuous on [a,b] and f(a) is less than K which is less than f(b), then K = f(c) for some number c in (a,b).
Proof:
(1) Let us start by defining three sequences {ai}, {bi}, and {Ii}
a1 = a
b1 = b
mi = midpoint of [ai,bi]
we can assume that f(mi) ≠ K since if it did, then c = mi and we are done with this proof.
if K is less than f(mi), then:
ai+1 = ai
bi+1 = mi
Otherwise K is greater than f(mi) and:
ai+1 = mi
bi+1 = bi
In this way, we can also define a sequence of nested intervals such that:
Ii = [ai,bi] and for all i, K lies between f(ai) and f(bi)
(2) Using the Nested Intervals Property of Real Numbers (see Lemma 1, here), we know that there exists a point c in I1 such that {c} = I1 ∩ I2 ∩ I3 ∩ ...
(3) We can see that both {ai} and {bi} have this point c as their limit. [See Definition 1, here for a definition of a limit]
(4) Since the function f(x) is continuous on (a,b), we know (see Definition 1, here for definition of continuous on a point) that:
for any positive real ε, there exists a positive real δ such that:
if abs(x-c) is less than δ, then abs(f(x) - f(c)) is less than ε
(5) By the definition of limits of a function (see Definition 1, here), it follows that the limit of {f(bn)} and the limit of {f(an)} is f(c).
(6) Since for all n, f(bn) is greater than K, it follows that f(c) ≥ K.
(7) Likewise for all n, f(an) is less than K so that f(c) ≤ K.
(8) But the only way that step #6 and step #7 can both be true is if f(c)=K.
QED
References
- Edwards & Penney, Calculus and Analytic Geometry
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