78 3.4 Properties of Continuous Functions 3.3.12 Prove Theorem 3.3.4. 3.3.13 ▷Consider the Thomae function defined on(0,1] by f (x) = 1 q , if x = p q , p,q∈N,where pandqhave no common factors; 0, if x is irrational. (a) Prove that for every ε >0, the set Aε ={x ∈(0,1] : f (x) ≥ε} is finite. (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∈Randa<b. Definition 3.4.1 We say that the function f : D→Ris bounded if there exists M∈Rsuch that | f (x)| ≤M, for all x ∈D. Theorem 3.4.1 — Boundedness Theorem. Let f : [a,b] →Rbe 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 ∈Rsuch 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. □ Definition 3.4.2 Let f : D→R. (i) f has anabsolute maximumat x0 ∈Dif f (x) ≤f (x0) for every x ∈D. (ii) f has anabsolute minimumat x0 ∈Dif f (x) ≥f (x0) for every x ∈D. Theorem 3.4.2 — Extreme Value Theorem. Let f : [a,b] →Rcontinuous on [a,b]. Then f has an absolute maximum and an absolute minimum on [a,b]. That is, there exists xM andxm 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 Ais a non-empty subset of Rand it is bounded (by the boundedness theorem). Applying the completeness axiom, Ahas a supremum, sayα=supA.
RkJQdWJsaXNoZXIy NTc4NTAz