78 3.4 Properties of Continuous Functions 3.3.12 Prove Theorem 3.3.4. 3.3.13 ▷ Consider the Thomae function defned on (0,1] by 1 p , if x = , p,q ∈ N,where p and q have no common factors; f (x)= q q 0, if x is irrational. (a) Prove that for every ε > 0, the set Aε = {x ∈ (0,1] : f (x) ≥ ε} is fnite. (b) Prove that f is continuous at every irrational point, and discontinuous at every rational point. 3.4 Properties of Continuous Functions In this section we present the most fundamental theorems about continuous functions on the real line. Wherever we use the interval [a,b] it is understood that a,b ∈ R and a < b. Defnition 3.4.1 We say that the function f : D → R is bounded if there exists M ∈ R such that | f (x)|≤ M, for all x ∈ D. Theorem 3.4.1 — Boundedness Theorem. Let f : [a,b] → R be continuous on [a,b]. Then f is bounded. Proof: Suppose, by way of contradiction, that f is not bounded. Then, for each n ∈ N, there is xn ∈ [a,b] such that | f (xn)| > n. The sequence {xn} is contained in [a,b], so it is bounded. It follows from the Bolzano-Weierstrass theorem (Theorem 2.4.1) that there exists a convergent subsequence, say, {xnk }, and x0 ∈ R such that limk →∞ xnk = x0. Note that by the comparison theorem (Theorem 2.1.2), x0 ∈ [a,b]. Using now the fact that f is continuous on [a,b], we conclude that limk →∞ f (xnk )= f (x0). Hence { f (xnk )} is bounded since it is a convergent sequence. This is a contradiction since | f (xnk )| > nk ≥ k for all k ∈ N. We have thus proved that f is bounded. □ Defnition 3.4.2 Let f : D → R. (i) f has an absolute maximum at x0 ∈ D if f (x) ≤ f (x0) for every x ∈ D. (ii) f has an absolute minimum at x0 ∈ D if f (x) ≥ f (x0) for every x ∈ D. Theorem 3.4.2 — Extreme Value Theorem. Let f : [a,b] → R continuous on [a,b]. Then f has an absolute maximum and an absolute minimum on [a,b]. That is, there exists xM and xm in [a,b] such that f (xm) ≤ f (x) ≤ f (xM), for all x ∈ [a,b]. Proof: Consider A = f ([a,b]) = { f (x) : x ∈ [a,b]}. Note that f (a) ∈ A, so A is a non-empty subset of R and it is bounded (by the boundedness theorem). Applying the completeness axiom, A has a supremum, say α = supA.
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