Feynman LiangMATH 355

Assignment #711/4/2013

Abbott: 4.3.2, 4.3.6(a), 4.3.7, 4.4.4, 4.4.6, 4.4.8, 4.4.13(a)

4.3.2 Given f : A→ R, g : B→ R, f (A) ⊆ B s.t. g ◦ f (x) is well defined on A.

f cts at c ∈ A, g cts at f (c) ∈ B =⇒ g ◦ f cts at c.

(a) Supply a proof using ε-δ characterization of continuity.

(b) Give another proof using sequential characterization of continuity (Theorem 4.3.2(iv)).

(a) If c is an isolated point, then ∃Nδ(c) : Nδ(c) ∩ A = {c}. Then since c is the onlyelement in Nδ(c), ∀x ∈ Nδ(c), g ◦ f (x) = g ◦ f (c) ∈ Nε(g ◦ f (c)) for any ε > 0, showingg ◦ f cts (Thm 4.3.2). Thus, assume c is a limit point. Since f cts at c, c ∈ A.

Let ε > 0 be arbitrary.Since g cts at f (c), ∃ξ0 > 0 s.t. | f (x)− f (c)| < ξ0 =⇒ |g ◦ f (x)− g ◦ f (c)| < ε.Since f cts at c, ∃δ0 > 0 s.t. |x− c| < δ0 =⇒ | f (x)− f (c)| < ξ0.Taken together:

|x− c| < δ0 =⇒ | f (x)− f (c)| < ξ0 =⇒ |g ◦ f (x)− g ◦ f (c)| < ε

So x ∈ Nδ0(c) =⇒ g ◦ f (x) ∈ Nε(g ◦ f (c)). Since ε arbitrary, by Theorem 4.3.2 g ◦ f cts. �

(b) If c is an isolated point, then for any sequence (xn) ⊆ A converging to c, ∃N0 ∈ N :∀n ≥ N0, xn = c. This implies ∀n ≥ N0, g ◦ f (xn) = g ◦ f (c) and therefore (g ◦ f (xn)) →g ◦ f (c), proving g ◦ f cts (Thm 4.3.2). Thus, assume c is a limit point.

Let (xn) be an arbitrary sequence s.t. (xn)→ c, c ∈ A, f cts at c.Since f cts at c , (xn)→ c =⇒ ( f (xn))→ f (c).Since g cts at f (c) and ( f (xn)) ∈ f (A) ⊆ B, ( f (xn)) → f (c) =⇒ (g ◦ f (xn)) →

g ◦ f (c).Taken together:

(xn)→ c =⇒ ( f (xn))→ f (c) =⇒ (g ◦ f (xn))→ g ◦ f (c)

So (xn) → c =⇒ (g ◦ f (xn)) → g ◦ f (c). Since (xn) and c arbitrary, by Theorem 4.3.2g ◦ f cts. �

4.3.6(a) Prove that Dirichlet’s function:

g(x) =

{1 if x ∈ Q

0 if x 6∈ Q

is nowhere continuous.

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Let x ∈ R arbitrary.Case I: x 6∈ Q

Since Q is dense in R, ∀ε > 0, Nε(x) ∩Q 6= ∅. Thus, we can produce a sequence whichconverges to x by letting n ∈ N be arbitrary and choosing any arbitrary xn ∈ N1/n(x),which is non-empty.

Notice that (xn) → x and g(x) = 0. However, ∀n ∈ N, xn ∈ Q so g(xn) → 1. Thus byCorrolary 4.3.3, g is not cts at x. Since x ∈ R arbitrary, g is nowhere cts.Case II: x ∈ Q

Since R \Q is dense in R, ∀ε > 0, Nε(x)∩ (R \Q) 6= ∅. Thus, we can produce a sequencewhich converges to x by letting n ∈ N be arbitrary and choosing any arbitrary xn ∈N1/n(x).

Notice that (xn) → x and g(x) = 1. However, ∀n ∈ N, xn ∈ (R \Q) so g(xn) → 0.Thus by Corrolary 4.3.3, g is not cts at x. Since x ∈ R arbitrary, g is nowhere cts. �

4.3.7 Assume h : R→ R cts and let K = {x : h(x) = 0}. Show that K is closed.

If K = ∅ then it is closed so assume that K is non-empty.Let c be an arbitrary limit point of K, which implies ∃(xn) ⊆ (K \ {c}) : (xn) → c.

Then since h cts, (h(xn))→ h(c). Furthermore, notice that (xn) ⊆ K so h(xn) = 0, ∀n ∈N.Thus, (h(xn)) = (0, 0, 0, . . .) and (h(xn))→ 0, implying h(c) = 0.

Since h(c) = 0, then c ∈ K. Since c arbitrary, K contains all its limit points and isclosed. �

4.4.4 Prove

f cts on [a, b] , ∀x ∈ [a, b] : f (x) > 0 =⇒ 1f

bdd on [a, b]

[a, b] is closed and bounded, so by HBT it is compact. Since f is cts on [a, b], EVT (Theorem4.4.3) implies that that:

∃x0 ∈ [a, b] : ∀x ∈ [a, b], f (x0) ≤ f (x)

Since f (x) > 0 for ∀x ∈ [a, b], f (x0) > 0 as well. Inverting the inequality yields∀x ∈ [a, b], 1

f (x0)≥ 1

f (x) , showing 1f bdd. �

4.4.6 Give an example or show impossibility:

(a) f : (0, 1)→ R cts and (xn) Cauchy s.t. f (xn) is not Cauchy.

(b) f : [0, 1]→ R cts and (xn) Cauchy s.t. f (xn) is not Cauchy.

(c) f : [0, ∞)→ R cts and (xn) Cauchy s.t. f (xn) is not Cauchy.

(d) f : (0, 1)→ R cts, bdd that attains a maximum value but not a minimum on (0, 1).

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(a) Let f : x 7→ 1x , xn = 1

n . Then f is cts on (0, 1) and (xn) → 0 ⇐⇒ (xn) Cauchy.However, f (xn) = n so ( f (xn)) diverges ⇐⇒ ( f (xn)) not Cauchy. �

(b) Impossible. (xn) Cauchy ⇐⇒ (xn)→ x. (xn) ⊆ [0, 1] and [0, 1] closed, so x ∈ [0, 1].Since f cts in [0, 1] and x ∈ [0, 1], Theorem 4.3.2 implies ( f (xn))→ f (x). Since ( f (xn))

converges, it is Cauchy. Thus, ( f (xn)) is Cauchy for all (xn) Cauchy. �

(c) Impossible. Same proof as (b), only changing domain from [0, 1] to [0, ∞) which is stillclosed and hence if (xn) → x is a Cauchy sequence in [0, ∞) than x ∈ [0, ∞). The rest ofthe proof is identical. �

(d) Let f : x 7→ x(1 − x). Then the maximum occurs when 0 = 1 − 2x =⇒ x =12 , yielding a maximum value for f of f (1/2) = 1

4 . However, any proposed minimumm0 = f (x0) is not minimal since if m0 ≥ 0.5 then f (x0 +

1−x02 ) is smaller and similarily if

m0 < 0.5 then f (x0 − 1−x02 ) is smaller. �

4.4.8

(a) Assume f : [0, ∞)→ R cts on its domain. Show

∃b > 0 : f uniformly cts on [b, ∞) =⇒ f uniformly cts on [0, ∞)

(b) Prove f : x 7→√

x is uniformly cts on [0, ∞).

(a) Let ε > 0 be arbitrary. [0, b] is compact (HBT) and so by Theorem 4.4.8 f is uniformlycts on [0, b]. f uniformly cts on [0, b] implies:

∃δ1 > 0 : ∀c ∈ [0, b], x ∈ Nδ1(c) =⇒ f (x) ∈ Nε( f (c))

Similarly, f uniformly cts on [b, ∞) implies:

∃δ2 > 0 : ∀c ∈ [0, ∞), x ∈ Nδ2(c) =⇒ f (x) ∈ Nε( f (c))

Choosing δ = max{δ1, δ2}, we get:

∀c ∈ [0, ∞) = [0, b] ∪ [b, ∞), x ∈ Nδ(c) =⇒ f (x) ∈ Nε( f (c))

Since ε > 0 arbitrary, f is uniformly cts on [0, ∞). �

(b) Consider [1, ∞). Let ε > 0 and x, c ≥ 1 be arbitrary. Choose δ = ε. Then ∀x : |x− c| <δ:

| f (x)− f (c)| =∣∣√x−

√c∣∣ = ∣∣∣∣∣

√x +√

y√

x +√

y(√

x−√y)

∣∣∣∣∣=

∣∣∣∣∣ x− y√x +√

y

∣∣∣∣∣< |x− y|= δ = ε

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Since ε > 0 and x, c ∈ [1, ∞) arbitrary, f is uniformly cts on [1, ∞). By part (a), f isuniformly cts on [0, ∞).

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4.4.13(a) Show that if f : A → R is uniformly cts and (an) ⊆ A Cauchy, then f (an) isCauchy.

Let f : A → R be uniformly cts on A and (xn) ⊆ A an arbitrary Cauchy sequence. Letε > 0 be arbitrary.

f uniform cts means ∃δ > 0 : ∀x, y ∈ A, |x− y| < δ =⇒ | f (x)− f (y)| < ε.an Cauchy means ∃N0 ∈ N : ∀m > n ≥ N0, |am − an| < δ. Letting x = am, y = an in

the uniform cts statement above, since (an) ⊆ A we have:

∀m > n ≥ N0, | f (am)− f (am)| < ε

Since ε > 0 arbitrary, ( f (an)) is Cauchy.�

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