THÔNG TIN TÀI LIỆU
Chapter Setting the Stage 1.1 Euclidean Spaces and Vectors ✁✂ ☎✄✝✆ ✞✠✟☛✡✌☞✎✍✑✏✓✒✔✟☛✡✕☞✎✍✑✏✓✒✔✟☛✡✖✏✗✟✘✄✚✙✜✛ ✞ ✢✂ ✣✄✝✆ ☞✎✍✤✙✜✒✔✟☛✡✥✙✠✟☛✡✕✏✗✟✦✄✚✞ ✁★✧✩✢✪✄✚✞✫☞✎✍✤✙✜✒✬✡✌☞✎✍✑✏✓✒✎✙✭✡ ☞✎✍✑✏✓✒✮✏✂✡✕✏✯✧✱✰✲✄✳✍✤✴ ✄✕✶✠✷✹✸✗✸✻✺✜✼✩☞✎✍✤✴✜✽✠✞✘✧✩✙✜✛ , ✞✾✒✿✄✕✶✠✷✹✸✗✸✻✺✜✼✩☞✎✍❀✛ ✞☛✽✠✙✜✒❁✄✕❂✠❃❄✽✠❅ , ,✵ ✟ ✟ ✟ ✁❇❆❈✢✂ ✄❉☞❊✁❋❆✖✢●✒✂✧❍☞❊✁❇❆❈✢■✒✲✄❏ ✁❑ ❆✖✙▲✁✪✧▼✢◆✡❖ ✢✂ Taking the plus sign gives (a); adding these ✁■●✡❈✧✗✧✗✧▲✡◗✁❙❘✫ ✟ ✄✌❚ ❯✮❘ ❱ ✁ ❯ ✟ ✡❲✙❁❚ ❨❳❬❩❪❭ ❯ ❳ ❘ ✁❫❩❴✧✗✁ ❯ The Pythagorean theorem follows immediately ✟ ☞❊❛❜✒❝✄❞ ❡❢✍◗❛✎❣❤ ☞❊❛❜✒ With ❵ proof, ❡❢✄✕❛✎❣ as in the❛❝✐❦ ❥ equality holds precisely when the minimum value of ❵ ❡ is 0, that for some Thus equality holds in Cauchy’s inequality precisely when and ❣ is, when are linearly dependent ❡❧✧✗❣◆✄❖ ❡✬ ✗ ❣❤ ❡ ❣ The triangle inequality is an equality precisely❣ when , that is, when the angle from to ❡ is 0, or when is a positive scalar multiple of or vice versa ❡✬ ✜✄♠ ♥☞♦❡♣✍✪❣■✒❫✡✥❣❤ ✣qr ❡s✍✪❣✯ ✗✡✌ ❣❤ ❡✬ ▲✍t ❣✯ ✣q✚ ❡✉✍✪❣❤ ❣❤ ▼✍t ❡✬ ✈q✚ ❡♣✍✇❣❤ , so Likewise, ❡s✧✓❣◆✄✕✰ ❡♣①✖❣ ❡★②❦❣❤ ▼✄♠ ❡● ✗ ❣❤ ❡❦②★❣◆✄✌③ ❡♣✄✕③ ❣◆✄✌③ (a) If then , so ; hence if also then or ❡❢✧▼④❦✄✝❣✪✧✠④ ❡◆②◆④❦✄✝❣⑤②◆④ ☞♦❡❢✍✥❣■✒☛✧✠④❋✄✝✰ ☞♦❡⑥✍❲❣■✒❀②◆④❦✄⑦③ (b) then and , so by (a), either ❡s✍✪If❣✪✄✕③ ④⑧✄✕③ and or ; the latter possibility is excluded ❡❋②★❡♣✄⑨③ ❡ ❣ ❡❦②★❣✪✄✕③ ❡★②★❣ (c) We always have If then too If not, then is a ❡ and ❡❦②⑩☞♦are ❡❦②★proportional, ❣✿✒✭✄✕ ❶ ③ identities with the plus and minus signs gives (b) nonzero vector perpendicular to , so This follows from the definitions by a simple calculation 1.2 Subsets of Euclidean Space (a)–(d): See the answers in the back of the text ✄✌❼ ✄ ✄ ☞❊➁➃➂❜✰✾✒✯➄❬✍✑✏✑q✥➁❢q✌✏▲➅ ❷ ❷❦❿❋➀ (e) ❷■❸❺❹❜❻ and ❽❾❷ ✄ ☞♦✰✣➂❜✰✾✒✹➅ ✄ ☞❊➇❄➂❨➁❬✒✯➄✈➇ ✟ ✡✥➁ ✟ q✚✏▲➅ (f) ❷ ❸❺❹❜❻ ✍✑❷❦ , and ✏✠➂❜➆❀✰▲➀ ➉■② ✰✈➅ , ❷ ➀ ➀ segment ➈ ✄✌❼ ✄ ✄ ✰✣➂✱✏✮➉■② ✰✣➂✱✏✮➉ (g) ❷ ❸❺❹❜❻ and ❽❾❷ ❷ ➈ ➈ Full file at ❽❴❷ is the union of the unit circle and the line Full file at Chapter Setting the Stage ✁✖✐ ✄ ☞✂✁✩➂❨✁✬✒ ❷■❸❺❹❜❻ , there is a ball If contained in ✁ ❷ is open, so every point of is an interior point of and hence of ❷ , so in fact ☎✄⑨❷ ❸❺❹✹❻ and is an interior ✒ ❷ ❸❺❹❜❻ is open by ☞ ✒ point of ❷ ❸❺❹✹☞ ❻ Thus Proposition 1.4a Next, ❷ and ❽❴❷ are the complements of ❷✝✆ ❸❺❹✹❻ and ❷■❸❺❹❜❻■❿ ❷✞✆ ❸❺❹✹❻ , respectively, so they are closed by Proposition 1.4b ✁✕✐ ✁ ✟ ❷ ❿❇❷ ✟ , some We use Proposition If centered is contained in either ✟ 1.4a ✁ ball ✁⑩✐ ✠at ✟ ❷ ✟ or ❷ and ✟ ✟ hence point of ❷ ❿ ❷ If ❷ ❷ , there are balls and centered ✁ in ❷ ❿ ❷ , so is an interior ✟ , respectively; the smaller of these balls is contained in ❷ ✟ ❷ ✟ , so again at and contained in and ❷ ❷ ✡✟ ✟ ✁ ❷ is an interior point of ❷ The complements of ❷ ❿❦❷ ✟ and ❷ by Exercise and Proposition 1.4b ✟ ❷ ✟ are ❷ ✆ ✟ ❷ ✟✆ and This 1.4: ✄ remarks preceding✄ Proposition ☞ ✒ ☞ ✒ follows from the ❷✞✆ ❸❺❹✹❻❫❿⑥❽❴❷ ❷✞✆ ❸❺❹❜❻ , whereas ❷ ❷■❸❺❹❜❻❫❿★❽❾❷ and ❷ ✆ One example (in ❥✔☛ and ❼ ❥ ❷ ✆ ❦ ❿ ❷ ✟ ✆ , respectively, which are both open ❥☞☛ is the disjoint union of ✁★✍✪❡● ✏✕✖✁ then ✁❑ ✜✄♠ ♥☞❊✁❦✍✪❡❾✒❫✡❲❡✬ ✈q✖✁✤✡✌ ❡✬ 1.3 Limits and Continuity Thus, if ❷✗✄✘ ☞✂✁▲➂❜❡➃✒ then ❷✗✄✘ ☞✂✁✤✡✌ ❡✬ ➂❾③➃✒ ☞♦✰✣➂❨➁❬✒❁✄✳✏ ➁✚✙ ✰ ☞♦✰✣➂❨➁❬✒❁✄✳✍✑✏ ➁✚✕ ✰ for and ❵ for ➇✧✣ ✰ ☞❊➇✬➂❜✰✾✒❁✄❈➇✛✎✢✜✤✣✦✥ as ☞❊❛✮➂❨❛❨✒✿✄ ✏✓✩✽ ★▲✫❛ ✪✬✣✦✥ ❛✭✣ ✰ as ✟ ✟ ➇❍➁❙ ✈q ✟ ☞❊➇ ✡◗➁ ✒ ☞❊➇✬➂❨➁✫✒✗ ✣q ☞❊➇ ✟ ✡◗➁ ✟ ✒✔✣ ✰ ➇❄➂❨➁✮✣ ✰ ✪ (a) Since , we have ❵ as ✞▲➇ ✪ ✍✇➁ ✪ ✣q⑤✞✫☞❊➇ ✪ ✡❲➁ ✪ ✒ ☞❊➇✬➂❨➁✫✒✗ ✫q⑤✞❬ ➇■ ✯✣ ✰ ➇❄➂❨➁✮✣ ✰ (b) Since , we have ❵ as ☞❊➇❄➂❨➁❬✰✒ ✣❏➁ ✚ ➇ ✣ ✰ ☞♦✰✣➂❨➁❬✒❁✄❈➁ as , so take ❵ ❵ ☞❊➇❄✲➂ ✱✫✒ ☞ ✱➃➂❨➁❬✒ ✳ ✱❇✄✖ ❶ ✰ ☞♦✰✣➂❜✰✾✒ ☞❊➇❄➂❜✰✾✒☛✄ ❵ ☞♦✰✣➂❨➁✫✒☛and ❵ are continuous for since ❵ is continuous except at Moreover, ❵ ✄✖✰ ➇✬➂❨➁ for all , also continuous ❵ ➁❢✄ ✰ ➁⑥✄✚➇ ✟ ➇t✄✳ ❶ ✰ The two formulas for ❵ agree along the curves except at ☞♦✰✣➂❜✰✾✒❁✄✖✰ and ☞❊➇✬➂ ✟ ➇ ✟ ,✒❁✄ ✟ ✣ ❶ , so✰ ❵ is➇✚continuous ✣ ✰ the origin It is discontinuous there since ❵ but ❵ as ☞❊➇❫✒✗ ✣q✚ ➇✿ ➇ ☞❊➇❫✔✒ ✣ ✰✲✄ ☞♦✰✾✒ ✧ ➇ ✣ ✰ ✱❇✄❈ ❶ ✰ ✱ Since for all , we have ❵ as ➇❴Suppose If is irrational, then ☞✳✱✣✒☛✄✴❵ ✱❋✄✕ ❶ ✰ ➇ ❵ ✱ ❊ ☞ ✿ ✒ ✖ ✄ ✰ ✱ ☞✳✱✫✒✿✄✖ ✰ , but there are points arbitrarily close to with If is rational, then , ❵ ❵ ❵ ➇ ✱ ☞❊➇❴✒✗✏ ✙ ✟ ✱❴ ✱ In both cases ❵ is discontinuous at but there are points arbitrarily close to with ❵ ☞❊➇❫✒✗ ✫q ➇■ ➇ ✱✇✄✕ ❶ ✰ ☞ ✱✣✒✑✄✕ ✳ ❶ ✰ Clearly➇ ❵ for all ✱ , so ❵ is☞❊➇❴continuous at If is rational, then✱ ❵ , but there are ✒✿✄✕✰ ✱ ❵ ❵ points arbitrarily close to with ; hence is discontinuous at If is irrational ✱ q✷✶ ☞❊➇❴✒✗ ✸✕✌✏✓✽✩✶ and ➇✂✍✹✱❴✵ ✏is✕ the distance from to the nearest rational number with denominator , then ❵ for ✺ ✵ ; ✱ q ✶ in any hence ❵ is continuous at (There are only finitely many rational numbers with denominator (a) ❵ (b) ❵ (c) ❵ , and ✄ ✰✣➂✬✏❝✍✍✌✏✎ ➉ ✄ ✰✣➂✱✏✓✒ ❯ ) is ❷ , for which ✑✓✒ ❷ ❯ ➈ ➈ The sets in Exercise 1a and 1f are both examples If ❷ ❸❺❹✹❻ , ❽❴❷ bounded interval.) Full file at Full file at 1.4 Sequences ❡✇✐✇❥ ⑨✰ ✄ ✁ ✙ ☛ ☞ ➂✄✂☎☞♦❡➃✒❨✒ ✄ ✁ ✁✥➄✆✂☎☞❊✁✬✒✘✐ ✁ ➅ , let Then☞ ➂❜❡➃is✒ open, and hence so is ☎ Given ❡t✐ and ✙✝✰ ✂☎☞❊➀ ✁✬✒❤✍✞✂✈☞♦❡➃✒✗ ✰✕ We have ✁s✍❋☎ ❡✬, so there✂ exists ✵ such❡ that ✵ ✄✝☎ But this says that ✕ ✵ , so whenever is continuous at One can replace “open” by “closed” in the hypothesis by the reasoning of the second paragraph of the proof of Theorem 1.13 ✂ points✒❧of The fact that since is a one-to-one correspondence between the ✂✈☞ ✁ ✄ ✁ ➆✠✟ the consequences ✁✪➄☛following ✂✈☞❊✁✬✒❤✐ ➅✭✄☞✂ ✎ ☞ ✒ that we shall use: (i) If ✟ ✄ , ➀ ❣ ✐✌✂☎☞ ☞ ➂❜❣✿✒❨✒ Suppose ✂ ✎ ☞ ✒ ✰ ✁ and the points of ☎ has ✂✈☞ ✒ ☎✳➆ ✟ (ii) If ✦✄✡☎ , ☞ ➂❜❣■✒ ✂ small enough so that ✄✍☎ Since , and let be is continuous, ✂ ✎ ☞♦❣✿✒ is a neighborhood of by Theorem☞ 1.13 it Hence it ➂❜❣✿✒ and the remarks following ✂✈☞ ✒ ❷ ❷ ❷ and points not in , and therefore contains points in and points not in contains points in ✂☎☞ ✒ ❣❲✐ ☞✎✂☎☞ ✒❨✒ ❷ It follows that ❽ ❷ ❽❴❷ ✙ ❣❲✐ suppose ✂☎❽ ☞ Conversely, ✂ ✎ Since is continuous, Hence it contains points❡★ in✐ not in ❷ It follows that ☞✎✂☎☞ ✒❨✒ ❡ ✄✏✂ ✎ ☞♦❣✿✒ ✙ ✰ ☞ ➂❜❡➃✒ ✁ ☞ ➂❜❡➃✒❨;✒ let be small enough ✄ ☞ ❷ ➂❜❡➃✒❨,✒☛and ✄ ☞✎let ✂ ✎ ✒ ✎ ☞ ❣ so that is a neighborhood of by Theorem 1.13 again ✂☎☞ ✒ ✂☎☞ ✒ ☞ ➂❜❡➃✒ ❷ and points❣◗not in ❷ , and so contains points in ❷ and points ✐✑✂✈☞ ✒ ❽❴❷ and hence ❽❾❷ 1.4 Sequences (a) Divide top and bottom by ✛ ✶ ✙ ✡✖✶ ✎ ➇❾❘✘✄ ✛ ✤ ✙❝✖ ✡ ✶ ✎ ✓✒ ✟ to get ✣ ✛ ✙ ✙ ✔✼✄✔✖✕ ❬✶ ✽✩✶❙ ✫q✌✏✓✽✩✶ ✣ ✰ ➇❴❘ ✰ ✟ ✛ ✞ ✍ ✟ ✛ ✞ ✶ (c) Diverges since is , , and for infinitely many ➇➃❘✦✍✪✞❬ ✜✄✳✏✱✴✜✽✫ ✶✉✍⑩❂❬✏ ✕ ✶✧✙⑤❂❝✡✕✏✱✴ ✎ whenever ✏ ✙ ✶ ✍✥✏ ✏ ➇➃❘✘✄✳✏✯✧ ✙ ✧ ✞ ✧✗✧✗✧ ✶ ✄ ✶ ✣ ✰ ➇➃❘ ✣ ✱ ➁✜❘ ✣✘✗ ☞❊➇❾❘☎➂❨➁✜❘▲✒ ✣ ✳☞ ✱➃✙➂ ✗✻✒ (b) each If and , then By continuity of addition and multiplication (Theorem 1.10) and the sequential characterization of continuity (Theorem 1.15), the result follows ✈☞❊✁✬✒✰✣✛✚ ✂ ✁ ✣ ❡ ✙ ✰ ✙ ✰ ✂✈☞❊✁✬✒✗✍✜✚✎ ✸✕ ✰ ✚ ✕ ✁❑✍✭❡✬ ✕ whenever ✵ If ✁❄❘ ✣ ❡ as , for any there such that✶✗✙ ✁✬❘ exists ✍❲❡✬ ✕ ✵ ✂☎☞❊✁✬❘▼✒❑✍✣✚✎ ✡✕ If , there exists such that ✢ ✵ whenever ✢ , and hence On ✂✈☞❊✁✬✒s✣✤ ❶ ✚ ✁✖✣ ❡ ✙⑨✰ ✙✚✰ ✁ ✵ the other hand, if as , there exists such that for every there is an ✰✮✕✚ ✁❦✍⑩❡✬ ✕ ✂✈☞❊✁✬✒■✍✥✚❨ ✙ ✁❄❘ ✄✳✏✓✽✩✶ ✁✬❘ ✣ ❡ ✂☎☞❊✁❄❘✠✒✦✣✦ ❶ with ✚ ✵ but Let be such a point for ✵ Then but ✁❙❘✉✐ ✁❙❘❋✄r ❶ ❡ ✁❙❘ ✣ ❡ ✁❄❘☎➅ If ❷ , ✙ ✰ , and , then the sequence ➀ ☞ must ➂❜❡➃✒ assume ❡ infinitely many distinct values, and for ,❡ all but finitely many of them are in ; thus is ✶ an accumulation of ❷ ☞♦❡❙➂✱✏✓✽✩point ✶❬✒ Conversely, if is an accumulation point of ❷ , for each positive integer the ball contains ❡ ✁●❘ points of ❷ other than ; let be one ❡ ❡✇✐ ❡⑤✐ ✽ ❡ If is an accumulation point of ❷ , then 1.14 and Exercise If ❷ by Theorem ❷ and is not ❡ an accumulation point of ❷ , there is a neighborhood of that contains only finitely many points of ❡ ❡ ❷ If is less than the minimum distance from to any of these points (which not coincide with since ❡◗✐ ✽ ☞ ➂❜❡➃✒ ❡ ❡◗✐ ✽ ❷ ), is a neighborhood of that is disjoint from ❷ , and hence ❷ Full file at Full file at Chapter Setting the Stage 1.5 Completeness ✄ ☞✎✍ ✏✠➂❄✍✑✏✓✽✜✛ ✙✜✒ ☞✎✏✓✽▼✛ ✙✈➂❄✏✓✒ ✍✑✏ ✏ (a) ❷ , so the inf and sup are and ❿ (b) The supremum is the 0th element of the sequence; the infimum is the limit of the odd-numbered subsequence ✄ ❃❄✽✁ ✫➂ ✥ ✒ ❃❄✽✁ ✥ ➈ , so the inf and sup are and (c) ❷ ➇❴❘✦✄✖✼✙✔ ✕❙☞ ✶✣❃❄✽✠✞✜✒ (Exercise 1c in ✂ 1.4) ✱✪✄ ✰☎✄ ✱ ✱ ✟ ✱ ✜ ✄✆✄✆✄❑✐r☞♦✰✣➂✱✏✓✒ ❾➇ ❘★✄ ✰☎✄ ✱ ✱ ✟ ✄✆✄✆✄ ✱☎❘ ✏✗✰ ❘ If , let , considered as a fraction with denominator ➇❾❘✾➅ ✱ ✱⑥✄✳✰ ➇●❘❧✄⑦✏✓✽✩✶ ➀ Then is a subsequence of the given sequence that converges to For , take ; for ✱s✄✳✏ ➇❾❘✘✄ ☞ ✶❧✍ ✏✓✒❨✩✽ ✶ , take ✔✟✞❲➇❾❘✘✡ ✄ ✠ ✠❙✡ ✄ ✠✟ ✠❫✄✕✰ , then and hence or (a) If ✝ ✱❴ ✕✌✏ ✱ ✄✌❆❧✏ ✱❫✏ ✙✌✏ (b) The limit is zero if , if , and nonexistent (or infinite) if ➇❄✂✄ ✛ ✮ ✙ ✕❈✙ ➇➃❘ ✕✖✙ ➇❾✆❘ ☛ ✄ ✛ ✙✤✡◗➇❾❘ ✕ ✛ ✙❝✡✥✙⑧✄✌✙ ➇❴❘ ✕t✙ We have ➇❫If☞❘ ☛ ✄ ,✙✤then ➇➃By induction, for all ✶ ✡ ➃ ➇ ❘ ✙ ➃ ➇ ✯ ❘ ✡ ➃ ➇ ✲ ❘ ✄ ▲ ✙ ❾ ➇ ❘ ✮ ✙ ➃ ➇ ❀ ❘ ✱ ✧ ❘ ⑨ ✄ ➃ ➇ ❘ ✛ ✛ ✛ ✛ ➇➃❘☎This being the case, Thus the sequence ➅ ✠ ✠✤✄ ✛ ✙✤✌ ✡ ✠ ➀ is increasing and✠❙bounded above , hence ✠ ✟ ✄✡✠❬✡✥✙ ✄✕✙ ✠❄✄✳ ✍✑✏ by 2, so it converges to a limit We have ➇✬❘ ✙ ✰ ✶ One example is , and hence or The latter alternative is impossible since for all ❾➇ ❘ ✶ ✁▼❘ ✄⑨➇➃❘✆☛ ✽✩➇➃❘ ➇❾❘✆☛ ✟ ✄⑨➇➃❘✆☛ ✡❲➇❾❘ (a) Let ✁✩❘✆☛ be✄ the✏❤✡✚th☞✎✏✓term of the Fibonacci sequence, so Since , we ✽ ✁✓❘✜✒✤✄ ☞✂✁✓❘❀✡⑨✏✓✒❨✽ ✁✓❘ ➇❴❘✆☛ ✶ ✶⑧✡⑨✏ obtain by dividing through by Replacing by we get ✁✓❘✆☛ ✟ ✄ ✂☞ ✁✓❘✆☛ ✡✖✏✓✒❨✽ ✁✓❘✆☛ , and substituting in ✁▲❘✆☛ ✄ ☞✂✁✓❘✯ ✡✖✏✓✒❨✽ ✁✓❘ gives ✁✩❘✆☛ ✟ ✄ ☞ ✙ ✁✓❘❝✡❈✏✓✒❨✽✣☞✂✁✓❘✤✡✖✏✓✒ ☞❊➇❴✒❁✄♠☞ ✙▲➇❁✡❦✏✓✒❨✽✣☞❊➇☛✡★✏✓✒❤✄✖✙❄✍ ☞❊➇☛✡❦✏✓✒ ✎ ➇ ☞✎✍■✒❁✄✏✍ is ✁✩an increasing ✁✓function of☞✂✁✓❘▲,✒✹and (b) The function ❵ ❵ ☞✎✍■✒✦✄✒✍ ✁✩❘ ✕✑✍ ✁✩❘✆☛ ✄ ✂✁✓❘✠✒✹✕ ☞ ✎ ☞ ● ✍ ✦ ✒ ✒ ✄ ✍ ✍✙✑✍ ❘ ✆ ❘ ☛ ✄ ✙ ✟ ✟ ❵ ❵ ❵ Hence,✁✜⑧ if ✄ ✧ , ✁▲and if ✍ then ✏ ✕✓✍ then ✁ ✟ ✄ ✙ ❵ ✙✓✍ ❘ ✕✓ ✶ ✁✩❘ ✙✓✍ ✶ Since for odd and for even Next, ✁✓✆ ❘ ☛ ✟ ✍ ✁✓❘✘✄ ✂☞ ✁✓❘▲✒❄✍ and✁✓❘✦✄♠☞✎✏■✡ ✁✓❘✯✍ , ✁ it❘✟ ✒❨follows ✽✣✂☞ ✁✩❘❑✡t✏✓that ✒ ✶ ❵ , which by the hint is positive for odd and negative ✶ for even ✁ ✗➅ ✁ ➅ (c) By (b), ➀ ✟ ❯ ✎ is an increasing sequence and ➀ ✟ ❯ is a decreasing ✍ ✠✔ ✠✟ ✔☞ ✠ ✒❁✡ ✄ ✠ ❯ sequence, bounded above and ✍ below, respectively, by Their limits and both satisfy ❵ ❯ , and hence both are equal to ✁❄✖❘ ✕▼➅ ❡ ✙❈✰ ☞ ➂❜❡➃✒ ✁●✗❘ ✕ ✌ If ➀ converges , then ✁■ ❘ contains ✁❄❘✙for all ☞✎sufficiently large ✌⑧Conversely, ❡ to , and ✘❝✐ ✏✠➂❜❡➃✒ ✄✌✙✈➂✹✞✈➂✚ ✫➂✆✄✆✄✆if✄ every ball about contains infinitely many , we can pick , and then for , ✶ ✙✷✶ ❯ ✎ so that ✁❙✗❘ ✕✑✐ ☞ ✌ ✎ ➂❜❡➃✒ ; then ✁❄✖❘ ✕✬✣ ❡ we can pick ❯ ✁●❘☎➅ If ❷ is bounded and infinite, let ➀ be a sequence of distinct points ❡ of ❷ By Theorem 1.19, this set ✂ 1.4 its limit is an accumulation point of ❷ (At has a convergent subsequence, and by Exercise in ✁❄❘ ❡ most one can be equal to ; throw it out if necessary.) ➇❙❘ ✶ ✘✍ ✙✘✱ ✼✜✛✣✢ ➇❾❘✲➄ ◆➅ ✶✥✤✧✦ ✘✍ ✙✘✱ ✦ , then ➀ ➇❙❘ ✙✘✱❍✡ and hence If✔✟✞ there many for which ✼★✛✣✢❤are ➇➃❘✩infinitely ✤✘✱❫✍ ✶ ✼★✛✣✢ ➇❾for ❘ ➄ all ✶✪✤✧✦◆➅✲q✘✱❍✡ ✝ If there are only finitely many for which , then ➀ ✦ ✔✟✞⑤✼✜✛✣✢❤➇❫❘⑧q✘✱❁✡ ✱❢q ✔✟✞⑤✼✜✛✣✢✂➇❫❘✲q✘✱ Since is arbitrary, we have ✝ for sufficiently large, and hence ✝ ✱❧✄ ✔✟✞ ✼✜✛✫✢✂➇❴❘ and hence ✝ ➇❫❘✖✕▼➅ 10 We define subsequence ➇❾❘✗✕ a✙⑦ ✼★✛✣✢ ➇➃❘◆➄✔✶✬➀ ✤ so that ❘✖✕✜✰ ✘ ☛ ✍❈☞✎✏✓✽ ➀ ✌☎✒✬✕t➇➃❘✗✕ ✭ have ✔✟✞✣✔✖✕✣✱ ✝ Full file at ✶ ✕ ❍ ✄ ✏ ✏ recursively We take , and for , we choose ✔✟✞ ❯ ✼✜✛✣✢ ❯ ✎ ✂ r ✡ ▲ ✏ ⑧ ➅ ✌ ✍ ✎ ☞ ✓ ✏ ✽ ✌✈✒ ❯✎ Then,✔✟✞◗ with of ✝ , we ❘✗✕✚✰ ✘ ☛ ➇❴❘✖✕✘✭✯✄ ✮ ✔✟as✞ in the✄ definition ✔✟✞⑤✼✜✛✣✢✂➇➃❘ ✭ ✝ ✭✯✮ ✝ It follows that ✝ Similarly for ✶ ✌✖✙ ✶ ✙ ✶ Full file at 1.6 Compactness ➇❾❘✖✕✚✣ ⑦✰ ✱ ♣✍ ✙ ❖➇❴❘ ✱ ✕ ❧✡ ✕ ✱ ✶ for infinitely many It follows that , then for any✔✟✞✣ ✔✖✕✣✱☎➇❾❘✲we have 11 If✔✟✞ ✼★✛✣✢❤➇➃❘✩✤✘ ✱✑✍ q✘✱✭ ✡ ✄✖✰ ✝ and ✝ ; since can be arbitrarily small, the same is true with ➁ ✼✜✛✣✢ ✔✟✞ ✔✟✞ ➇❙❘✘✍ ❴✱ ➃q ✔ ✕✣✱ ✤ ✶ 12 With ✭ ✮ and ✮ as in the definition of and➁ ✝ ✤✴✱❧✍ , the assertion that for ✢ q✴✝ ✱✑✡ ✦ ✤ ✱ ✍ q and ✮ for ✢ If this holds, then is✔✟✞✣ equivalent to✔✟✞ the✼★✛✣assertion that ✭ ✮ ✔✖✕✣✱☎➇❾❘★q ✢❤➇➃❘★q ✱✲✡ ✔✟✞✣✔ ✕✣✱☎➇❴❘ ✄ ✱◆✄ ✔✟✞ ✼✜✛✣✢❤➇❾❘ if ✝ ✝ for every , and hence ✝ ✝ ✙⑨✰ ✱ ✍ q✕➁ q Conversely, q ✱ ✡ ✮ ✭✯✮ for the latter condition holds, then for any there exists such that ✦ ✤ ➇❾❘❀✍ ✱❴ ✈q ✶✥✤ ✔✟✞◗➇❾❘✘✄✴✱ , and so ; hence ✝ for ✁ ✁ ✁ 1.6 Compactness ✄✖❥ ☞❊➇❴✒✿✄✄✂✆☎ ,❵ ✄❈❥ ☞❊➇❴✒❁✄✖➇ ✟ ,❵ (b) One example is ❷ ✄ ☞♦✰✣➂✱✏✓✒ ☞❊➇❴✒✿✄ ✏✓✽✩➇ ♠ (a) One example is ❷ ,❵ ✄✞✝ ✄ ✝ ☞ ✒ ✟ ✄✟✝ ☞ ✒ ☞ ✒ ❷ compact ❵ ❷ compact ❵ ❷ ✄t❵ ❷ bounded (b) ❷ bounded ✁❑✱➂❨✁ ✟ ➂✆✄✆✄✆✄ If ❷ is compact and ☎ is an infinite subset of ❷ , let be a sequence of distinct points of ☎ ✚ ✚ This sequence has a convergent subsequence whose limit lies in ❷ , and is an accumulation point of ☎ (Exercise 6, ✂ 1.4) Conversely, suppose ❷ is not compact If ❷ is not closed, there is a sequence ✚✿✐ ✁❙❘☎➅ ✁■❘✜➅ ➀ ✁❙❘❬ ✣ in ✥ ❷ that converges to a point ✁❁✱➂❨✁ ❷ ✆ ✆➂ ,✄✆and if ❷ is not bounded, there is a sequence ➀ in ❷ with ✆ ✄ ✄ ➅ ✟ In either case, the set ➀ is an✚ infinite subset of ❷ with no accumulation point in ❷ (In the first case, the only accumulation point is ; in the second case, there is no accumulation point (a) One example is ❷ at all.) ✁✬❘✾➅ ☞❊✁❄❘▲✒ ✌ ✕ ✏✓✽✩✶ Some subsequence If not, there☞ ✚ ✒✿ is✄ a sequence ✔✟✞ ☞❊✁❙❘✗✕✩➀ ✒✿✄✖✰ in ❷ such that ❵ but then ❵ ✝ ❵ , contrary to assumption ✌✏ ✁■❘✲✐ ✶✪✤ ❘ ✁❄❘✲✐ ➀ ✁✬❘✖✕✠➅ has a limit ●✐ ✚ ✡✠ ✶ ❷ ; ✁●❘✖✕✜➅ By Bolzano-Weierstrass: For , pick✁ ✐ ❷ for all , so❘ some subsequence ➀ ❘ ✚❤✐ ❘ ❷ Then ✌ ✙ ✶ ✚ ✶ ✚❤✐ ❷ But since ❯ ❷ for ❷ ❷ converges to a point , is actually in for all , i.e., ✄ ✁ ❯ If the sets ✁ ❘ covered ❷ , By Heine-Borel: Let be an open ball containing ❷ , and let ❯ ⑨➆❝❷ ❶ ❼ ✁ ❘✘✄ ✁ But this is false since ❷❧➆ ✁ ✄ ❷ ✄✌ there would be a❘ finite subcover; that is, ❷ ✄ ❘✘✑ ✄ ❶ ❼ ✁ ✁ ❘ ✄✌ Thus the sets not cover ❷ , that is, ❷ ❷★➆ ✑ ☞☛ ✠ ✁ ✟ ☎ ➀ , there is a sequence 1.7 Connectedness (a) The two branches ( ➇ ✙ ✰ and ➇ ✕ ✢●❘✾➅ ✰ in ☎ that converges to , i.e., ) (b) One point in the set and the rest of the set (c) The intersections with the half-spaces Full file at ✁ ☛ ☛ ✁❝✍✘✢●❘❬ ✯✣ ✰ ☞ ✁ ➂ ✒☛✄✖✰ ; thus ✌ ☎ ☞ ✁ ➂ ✒❝✄✳✰ ✁✿❘ ✐ ✁ ❙✢ ❘❢✐ ✁ ☎ is closed, but ✌ ☎ (b) Then there exist , ✁❙❘✑Suppose ✍ ✢❙❘✣ ✣ ✰ is compact, ✁❁❘✮☎ ✣ such ✁ ✐ that t ✁ ✁ is ✢❄❘ Since ✣❏✁ ✁⑩compact, ✐ ✄ by passing to a✁ subsequence ✟ ✄✌❼ we may assume that ☎ ☎ , contradicting ☎ But then also , so ☞❊➇❄➂❨➁❬✒❤➄▼➁❢q ✰✈➅ ✄ ☞❊➇✬➂❨➁✫✒✯➄▼➁ ✤✍✂✆☎☎➅ ✁ ✄ ➀ ➀ (c) One example is ,☎ (a) If ✁⑩✐ ☛ ➇ ✙ ✰ and ➇ ✕ ✰ Full file at Chapter Setting the Stage ❡❫➂❜❣ ❡ ❣ ❣ are points in the unit sphere ❷ , the plane through , , and the ❡ origin ❣ (that is, the linear span of and ) intersects in ✄❖ a circle, ❡ ❣ ❷ ❣⑩ ✍❝❡ and either of the two arcs ❡ between and provides a continuous path in ❷ from to (If , any great circle through will do.) This argument works in any number of dimensions If ❡ ➇❄➂❨➁❾➂✁ ✐✄✂ strictly increasing (i) ❵ ✕ is✴ ☞❊➇❫✒ nor q strictly ☞❊➁❬✒ decreasing, ☞❊➁✫✒ ✤ one ☞✆ ☎✒ can find ☞❊➇❴✒ points ✤ ☞❊➁❬✒ ☞❊➁❬✒ such q ☞✆that ☎✒ ➁ neither ✕☎ ❵ ❵ ❵ ❵ ❵ ❵ ❵ , and (ii) either ❵ and , or and ; we ☞❊➇❫✒✭✄ ☞❊➁❬✒ ☞❊➁❬✒✭✄ ✆☞ ✈✒ assume the former alternative If ❵ ❵ or ❵ ❵ , then ❵ is not one-to-one Otherwise, ☞ ☞❊➇❴✒ ➂ ☞❊➁❬✒❨✒ ☞ ☞✆ ☎✒ ➂ ☞❊➁❬✒❨✒ the intervals ❵ and ❵ are nonempty, and ☞❨one in the other Assuming ❵ ❵ ☞❊➇❄➂❨➁❬is✒❨✞ ✒ contained ✝✳☞ ☞❊➇❫✒ ➂ ☞❊➁✫✒❨✒ ☞❨☞❊➁❾➂✁ ☎✒❨✒✞✝ ❵ ☞ is✆☞ ☎continuous, ❵ ❵ ❵ the intermediate value theorem implies that and ❵ ✒ ➂ ☞❊➁❬✒❨✒ ☞❊➇✬➂❨➁✫✒ ☞❊➁❾✁➂ ☎✒ , so there are points in and at which ❵ takes the same value, and again ❵ is not ❵ ❵ If ➇ one-to-one ✄ ✟ Suppose ❷ ❿★❷ ✟ is disconnected, ✂✄ ☞ ✟ ✁ so ✒ ❷ ☞ ❿❦ ✟ ❷ ✒ ❿ ☎ where neither nor ☎ intersects ✟ the closure ✟ of✁ the other one Then is a disconnection of ❷ unless either ❷ ❿ ❷ ☎ ☎ or ❷ ✁❷ ❷ ✁ is empty, i.e., ❷ ✄ or ❷ ✄ ☎ Likewise, we must have ❷ ✟ ✄ or ❷ ✟ ✄ ☎ It cannot be that ❷ and ✁ ✁ ✁ ✟ ❷ ✟ are both contained in (resp ☎ ), for then ☎ (resp ) would ✟ be✟ empty; ✄✌ ❶ ❼ so ❷ ✄ and ❷ ✄☞☎ or vice versa Either alternative contradicts the assumption that ❷ ❷ ✢✟ ✟ ✄✳✏ ✙✌✏ ❷ ❷ is connected when ✟ by Theorem 1.25, but not when ✟ For example, take ❷ to be the unit sphere (Exercise 2) and ❷ ✟ to be a line through the origin; the intersection consists of two points ✄ ✁ ✁ ✁ ✁✥✐ ✁ ✁ ✁ , there is✟ a ball centered at that Suppose ❷ ❿ ☎ where and ☎ are open and disjoint ✁✝✐ ✽ If ✁ ✁ ✄ ❼ ✁ ☎ Likewise ☎ is contained in and hence is disjoint from ☎ ; hence , so ❿ ☎ is a disconnection of ❷ ✄ ✁ ✁✕✐ ✁ , there is a ball centered Conversely, suppose ❷ is open and ❷ ❿ ☎ is a disconnection ✁ If ✁ at ✟ that✄ is❼ contained in ❷ (since ❷ is open) and a ball centered at that does not intersect ☎ (since ✁ ✁ ✁ ) The smaller of these two is a ball centered at that is contained in Thus every point ☎ ✁ ✁ ✁ of is an interior point of , so is open; likewise, ☎ is open ❷ ✄ ✁ ✄ ❿ ✁ ✁ where and ☎ are closed and disjoint, it is immediate that❡❋✐ ❿ ☎ is a disconnection of ✁ ✁ if ❷ is✟ closed and ❿ ☎ ❡ is ✐ ✽ a disconnection Since ❷ is closed, we ❷ Conversely, ❡★✐ ❡★✐ of✁ ❷ , suppose ✁ ✄✌❼ ✁ ❷ ; since ☎ ☎ Hence have , we have , so is closed Likewise, ☎ is closed If ☎ ☞❊✁✬✒❝✄ ✰ ✁t✐ ☞❊✁✬✒❀✄ ✏ ✁t✐ ❷ ❿❋❷ ✟ is a disconnection ❷ and ❵ ❷ ✟ Each If ❷ of ❷ , define ❵ for for ✟ ✁ ✁ ✟ point of ❷ has a neighborhood that does not intersect ❷ , so that ❵ is constant on ❷ ; likewise with ❷ and ❷ ✟ switched It follows that ❵ is continuous on ❷ ✰✣➂✱✏▲➅ ♣✄ ✎ ☞ ✰✈➅▲✒ ✟ ✄ ❵ ✎ ☞ ➀ ✏▲➅▲✒ If ✁♠✐ Conversely, if ❵ maps ❷ continuously onto ➀ , let ❷ ❵ ➀ and ❷ ✟ ✟ ✄⑨❼ ✟ ☞❊✁✬✒☛✄✕✰ ✁✕✐ ✽ ✟ since ❵ is continuous, so , and likewise with ❷ and ❷ ❷ , then ❵ ❷ Thus ❷ ❷ ❷ ✟ switched, so ❷ ❿★❷ ✟ is a disconnection of ❷ ✄ ☞ ✁ ✟ ✁ ✒ ☞ ✟ ❷ ✟ ❿ ☎ is a disconnection of ❷ Then ❷ ❿ ❷ Suppose ✟ ✁ or ❷ ❷ ☎ is empty The latter alternatives are impossible: If ❷ ✁ ✁ ✄ ✁ ✟ since does not intersect the closure of ☎ , we would have definition of disconnection ✁✪✐ ☞❊✁✬✒☛✄✖✰ ☞❊✁✬✒ ✙ ✰ ☎ ✒ is a✄ disconnection of ❷ unless ❼ ✁ , say, ✟ ✄⑨❼ then ❷✴✄ ☎ ; but ❷✘✄ ✁ ☎ , contrary to the ✟ ☞❊✁✬✒ ✕ ✰ ☞✎✍✯✁✬✒☞✕⑤✰ Pick☞✎✍✯✁✬✒✬❷ ✙✖ If✰ ✠ we are done Otherwise, either ✠ or ✠ , in which☞❊✢●case ✒✯✄✚✠ ✰ ✠ ✠ or respectively; either way, the intermediate value theorem implies that for some ✢ ✐ ❷ Full file at Full file at 1.8 Uniform Continuity ☞ ✫➂✱✍✑✏✓✒☛✄✕❂ ☞❊➇❄➂❨➁❬✒❤✐ ☞❊➇❄➂❨➁❬✒❁✄✖✰ ➇❢✄✖➁ ❷ such that ❵ , so there is a point , i.e., ➁✉✄✖✼✄✔ ✕❙☞♦❃❄✽✩➇❫✒ ✮ ✰ ✕⑤➇❇q⑤✙ 11 (a) The graph , , is arcwise connected almost by definition (it’s an arc!), and ❷ is its closure (Check that every point in ❷ ✆ has a neighborhood ✰✈➅❢that ② ✍✑does ✏✠➂✱✏✮➉ not intersect ❷ , and that every neighborhood contains points of the graph ➈ ➁✉✄❈✼✄✔ ✕❙☞♦❃❄✽✩➇❴✒ of every point on the vertical line segment ➀ ) So ❷ is connected by Exercise ✂⑤➄ ✰✣➂✱✏✮✓ ➉ ✣ ✂✈☞♦✰✾✒❦✄ ☞ ✙✈➂❜✰✾✒ ✂☎☞✎✏✓✒★✄ ☞♦✰✣➂✱✏✓✒ (b) Suppose and ✶ ❛✗❘ The ➈ ❷ is continuous and satisfies ✂ ✐ ✰✣first ➂✱✏✮➉ component of is continuous, so by the intermediate value theroem, for each there exists ❵ ➈ ✓☞❊❛✎❘✠✒❁✄ ✏✓✽✠✙ ✶ ✂☎☞❊❛✎❘▼✒❁✄ ☞✎✏✓✽✠✙ ✶❾➂❜✰✾✒ ➇ ✏✓✽✠✙ ✶ ❛ so that ❵ ❛❨❘ hence (= the only✏✓✽✠point ) As ❛ ❯ ✌❦✄ ❶ and ✶ ✙ ✶ in ❷ ✏✓✽✠with ✙ ✌ -coordinate goes from to ( ✍✑✏ ,) ❵ ✏ must assume all values between and , and hence ➇ ❵ ✟ must assume all values between ➁ and (again because there is only one ✍✑✏ point✏ in ❷ with a given -coordinate in range from to ) By passing this range, and the -coordinates of these points ❛ ❘ ✣ ✁❛ ❛ to a subsequence, by ✂ Bolzano-Weierstrass we may assume✍ that Every neighborhood of contains points at which ✟❵ assumes any given value between ✏ and ✏ , so ❵ ✟ cannot be continuous at ❛ , contrary to assumption 10 ❵ ☞✎✏✠➂✹✞✜✒❑✄✳✍✤✙ and ❵ 1.8 Uniform Continuity ✁❦✍✇✢✂ ✏✕✚☞ ✽☎✄ ✒ ✓ ✒✁✆ ✂☎☞❊✁✬✒■✍ ✂✈☞❊✢●✒✗ ✸✕ Given , if then ✆ ✆ ✆ ✆ ✆ ❛ ✕ ❛ ✎ ❛ ☞✳✱✘✡✣✗✻✒ ✍ ✱ ✕ ✗ ☞✳✱✘✡❲❛❨✒ ✎ (a) ✝✟✞ ✌ ✝✠✞ ✌ , so ➇❄➂❨➁ ➇✿ ✆ q✚☞❜ ➇⑧✍⑥➁❙ ❨✡⑤ ➁❙ ❺✒ ✆ q✚ ➇❧✍⑥➁❙ ✆ ✡ ➁❙ ✆ ➇ ➁ , and likewise with and switched; (b) For any ➇■ ✆ ✍t ➁❫we ✆ have q✚ ➇♣✍✇➁❙ ✆ hence ✡ ✡ ✡ ✡ ✂☎☞❊✁✬✒☛✍ ✂☎☞❊✢●✒✗ ✕ ✁✿➂❨✢⑤✐ ✁◆✍⑩✢❤ ✕ ✙✕✰ ✱ ➂ ✙✌✰ ✟ ❷ and ✒ Given , we so ☛●☞❊✁✬✒✦ ✍ ☛■can ☞ ☞❊✢●✒✗choose ✕ ✟ ✵ ✵ ✹ ✁✿➂❨that ✢ ✐ ✁✥✍✖✢✂ ✟ ✕ whenever ✄ ✞ ✔ ✕❙☞ ✱➂ ✟ ✵ ❷ and ✵ ✟ Let ✵ ✵ ♥☞✎✂❤and ✡✌☛❴✒✻☞❊✁✬✒❁✍⑤☞✎✂✂✡✍☛❫✒✻☞❊✢●✒✗ ❍q✚ whenever ✂☎☞❊✁✬✒✿✍ ✂☎☞❊✢●✒✗ ✩✡✌ ☛■☞❊✁✬✒✿✍✎☛■☞❊✢●✒✗ ✸✕ ✁✿➂❨✢◗✐ ✁★✵ ✍✇✢❤ ✏Then ✕ whenever ❷ and ✵ ✂ ✁●❘☎➅ ✙ ✰ ✙❏✰ Suppose continuous is Cauchy Given , there exists so that ✂☎☞❊✁✬✒✭✍☞✂☎☞❊✢■is✒✗ uniformly ✕ ✁✥✍❈✢✂and ✕➀ ✁ ❯ ✍✖ ✁❙❘✈ ✵ ✕ such that ✂✈☞❊✁❄❘▼✒✹➅ whenever ✢ ✵ whenever ✌▼➂ ✶ ✙ ✂☎☞❊✁ ❯ ✒❤✍ ✂✈☞❊✁❙❘✠✒✗ ✵ ,✕ and there exists ✌▼➂ ✶ ✙ ✢ It follows that whenever ✢ , so ➀ is Cauchy For the ➇❫❘✦✄ ✏✓✩✽ ✶ ☞❊➇❴✒✿✄✳✏✓✽✩➇ ☞❊➇❴✒☛✄❈✸✻✺✜✼✩☞♦❃❄✽✩➇❴✒ counterexample, take and ❵ or ❵ ✂✈☞ ✒ ✁■❘✜➅ ✂☎☞❊✁❄❘▼✒✗ ✣ ✥ If ❷ is unbounded, we can find a sequence ➀ If also ❷ is bounded, ✁■in❘✾➅ ❷ such that by passing ✁ to✍✉ a ✁❙subsequence we may assume that ➀ converges to some limit (which may not be in ❘✣ ✌ ✶ ✌ will be ✂☎☞❊as small as we please provided and are sufficiently large, but for any we ❷ ) Then✶✚✙ ❯ ✌ ✁✬❘✜✒✗✏ ✙✚ ✂✈☞❊✁ ❯ ✒✗ ✱✡✖✏ ✂☎☞❊➇ ❯ ✒●✍ ✂✈☞❊✁❙❘✠✒✗☎ ✤⑨✏ ✂ can find such that and hence Thus is not uniformly continuous on ❷ Full file at ✙ ✰
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