Math 521 Archives - Page 4 of 8 - Shawn Zhong

Math 521 – 3/23

$Sequences Approaching Infinity • Let {s_n } be a sequence of real numbers s.t. • ∀M∈R,∃N∈N s.t. s_n≥M,∀n≥N • Then we write s_n→+∞ • Similarly if ∀M∈R,∃N∈N s.t. s_n≤M,∀n≥N • Then we write s_n→−∞ Upper and Lower Limits • Definition ○ Let {s_n } be a sequence of real numbers ○ Let E be the set of x (in the extended real number system) s.t. ○ s_(n_k )→x for some subsequence {s_(n_k ) } ○ E contains all subsequential limits of {s_n } plus possibly +∞,−∞ ○ (lim⁡sup)_(n→∞)⁡〖s_n 〗=s^∗=sup⁡E is called the upper limit of {s_n } ○ (lim⁡inf)_(n→∞)⁡〖s_n 〗=s_∗=inf⁡E is called the lower limit of {s_n } • Example 1 ○ s_n=(−1)^n/(1+1/n)={−1/2,2/3,−3/4,4/5,−5/6,…} ○ (lim⁡sup)_(n→∞)⁡〖s_n 〗=sup⁡{−1,1}=1 ○ (lim⁡inf)_(n→∞)⁡〖s_n 〗=inf⁡{−1,1}=−1 • Example 2 ○ lim_(n→∞)⁡〖s_n 〗=s⇒(lim⁡sup)_(n→∞)⁡〖s_n 〗=(lim⁡inf)_(n→∞)⁡〖s_n 〗=s § All subsequential limits of a convergent sequence § converge to the same value as the sequence ○ (lim⁡sup)_(n→∞)⁡〖s_n 〗=(lim⁡inf)_(n→∞)⁡〖s_n 〗=s⇒lim_(n→∞)⁡〖s_n 〗=s § ⇒sup⁡E=inf⁡E § ⇒E={s} § ⇒ All subsequential limits = s § ⇒lim_(n→∞)⁡〖s_n 〗=s Theorem 3.17 • Let {s_n } be a sequence of real numbers, then • s^∗∈E ○ When s^∗=+∞ § E is not bounded above, so {s_n } is not bounded above § There is a subseqnence {〖s_n〗_k } s.t. 〖s_n〗_k→∞ § So s^∗=+∞∈E ○ When s^∗∈R § E is bounded above § And at least one subsequential limit exists i.e. E≠∅ § By Theorem 3.7, E is closed i.e. E=E ̅ § By Theorem 2.28, s^∗=sup⁡E∈E ̅ § Therefore s^∗∈E ○ When s^∗=−∞ § Then E={−∞} § s_n→−∞ and s^∗=−∞∈E • If x s^∗,then ∃N∈N s.t.s_n x for n≥N ○ If ∃x s^∗ with s_n≥x for infinitely many n∈N ○ Then ∃y∈E s.t. y≥x s^∗ ○ This contradicts the definition of s^∗ • Moreover s^∗ is the only number with these properties ○ Suppose p,q∈E,p≠q s.t. the property above holds for p,q ○ Without loss of generality, suppose p q ○ Choose x s.t. p x q ○ Since p satisfies the property above ○ ∃N∈N s.t. s_n x,∀n≥N ○ So no subsequence of {s_n } can converge to q ○ This contradicts the existence of q ○ Therefore only one number can have these properties$

Math 521 – 3/21

$Theorem 3.11 • Statement (a) ○ In any metric space X, every convergent sequence is a Cauchy sequence • Proof (a) ○ Suppose p_n→p ○ Let ε 0,then ∃N∈N s.t. d(p,p_n ) ε/2, ∀n≥N ○ d(p_n,p_m )≤d(p,p_n )+d(p,p_m ) ε/2+ε/2=ε, ∀n,m≥N ○ So {p_n } is a Cauchy sequence • Statement (b) ○ If X is a compact metric space and {p_n } is a Cahchy sequence ○ Then {p_n } converges to some point of X • Proof (b) ○ Let {p_n } be a Cauchy sequenec in compact metric space X ○ For N∈N, let E_N={p_N,p_(N+1),…} ○ By Theorem 3.10,lim_(N→∞)⁡〖diam (E_N ) ̅ 〗=lim_(N→∞)⁡〖diam E_N 〗=0 ○ By Theorem 2.35, (E_N ) ̅ as closed subset of X is compact ○ Since E_(N+1)⊂E_N, (E_(N+1) ) ̅⊂(E_N ) ̅,∀N∈N ○ By Theorem 3.10 (b), ∃!p∈X s.t. p∈(E_N ) ̅, ∀N∈N ○ Let ε 0 be given,∃N_0∈N s.t. diam (E_N ) ̅ ε,∀N≥N_0 ○ Since p∈(E_N ) ̅,d(p,q) ε,∀q∈E_N={p_N,p_(N+1),…}⊂(E_N ) ̅ ○ In other word, d(p,p_n ) ε for n≥N_0 ○ So lim_(n→∞)⁡〖p_n 〗=p • Statement (c) ○ In Rk, every Cauchy sequence converges • Proof (c) ○ Let {(x_n ) ⃗ } be a Cauchy sequence in Rk ○ Let E_N={(x_N ) ⃗,(x_(N+1) ) ⃗,…} ○ For some N∈N,diam E_N 1 ○ Then the range of {(x_n ) ⃗ } is {(x_1 ) ⃗,…,(x_(N−1) ) ⃗ }∪E_N ○ By Theorem 2.41, every bounded subset of Rk has compact closure in Rk ○ (c) follows from (b) Complete Metric Space • Definition ○ A metric space X is said to be complete if ○ every Cauchy sequence converges in X • Examples ○ Rk is complete ○ Compact metric space X is complete ○ Q is not complete (convergence may lie outside of Q) Monotonic Sequence • A sequence {s_n } of real numbers is said to be • monotonically increasing if s_n≤s_(n+1), ∀n∈N • monotonically decreasing if s_n≥s_(n+1), ∀n∈N • monotonic if {s_n } is either monotonically increasing or decreasing Theorem 3.14 (Monotone Convergence Theorem) • Statement ○ If {s_n } is monotonic, then {s_n } converges if and only if it is bounded • Proof ○ By Theorem 3.2 (c), converge implies boundedness ○ Without loss of generality, suppose {s_n } is monotonically increasing ○ Let E=range {s_n }, and s=sup⁡E, then s_n≤s,∀n∈N ○ Given ε 0,∃N∈N s.t. s−ε s_n≤s,∀n≥N ○ Since s−ε is not an upper bound of E, and {s_n } is increasing ○ s−s_n ε,∀n≥N⇒lim_(n→∞)⁡〖s_n 〗=s$

Math 521 – 3/19

$Cauchy Sequence • A sequence {p_n } in a metric space X is said to be Cauchy sequence • If ∀ε 0, ∃N∈N s.t. d(p_n,p_m ) ε,∀n,m≥N Diameter • Let E be a nonempty subset of metric space X • Let S be set of all real numbers of the form d(p,q) with p,q∈E • Then diam S≔sup⁡S is called the diameter of E (possibly ∞) • If {p_n } is a sequence in X and E={p_N,p_(N+1),…} • Then {p_n } is a Cauchy sequence if and only if (lim)_(N→∞)⁡〖diam E_N 〗=0 Theorem 3.10 • Statement (a) ○ If E ̅ is the closure of a set E in a metric space X, then diam E ̅=diam E • Proof (a) ○ diam E≤diam E ̅ § This is obvious since E⊂E ̅ ○ diam E ̅≤diam E § Let p,q∈E ̅ § Let ε 0, then ∃p^′,q^′∈E s.t. d(p,p′) ε/2, d(q,q′) ε/2 § d(p,q)≤diam E □ d(p,q)≤d(p,p^′ )+d(p^′,q^′ )+d(q^′,q) □ ε/2+d(p^′,q^′ )+ε/2 □ =ε+d(p^′,q^′ ) □ ≤ε+diam E □ Since ε 0 was arbitrary, d(p,q)≤diam E § So diam E ̅≤diam E ○ Therefore diam E ̅=diam E • Statement (b) ○ If K_n is a sequence of compact sets in X s.t. ○ K_n⊃K_(n+1),∀n and (lim)_(n→∞)⁡〖diam K_n 〗=0 ○ Then ⋂24_(n=1)^∞▒K_n consists of exactly one point • Proof (b) ○ Let K=⋂24_(n=1)^∞▒K_n ○ By Theorem 2.36, K is not empty ○ If K contains more than one point, diam K 0 ○ But K_n⊃K, ∀n∈N, then ○ diam K_n≥diam K 0⇒lim_(n→∞)⁡〖K_n 〗≥diam K 0 ○ This contradicts lim_(n→∞)⁡〖diam K_n 〗=0 ○ There can only be one point in K$

Math 521 – 3/16

$Theorem 2.41 (The Heine-Borel Theorem) • For a set E∈Rk, the following properties are equivalent (a) E is closed and bounded (b) E is compact (c) Every infinite subset of E has a limit point in E • Proof (a)⇒(b) ○ If (a) holds, then E⊂I for some k-cell ○ (b) follow from § Theorem 2.40 (I is compact) § Theorem 2.35 (Closed subsets of compact sets are compact) • Proof (b)⇒(c) ○ See Theorem 2.37 • Proof (c)⇒(a) ○ Suppose E is not bounded § Then E contains points (x_n ) ⃗ s.t. |(x_n ) ⃗ | n, ∀n∈N § {(x_n ) ⃗ } is an infinite subset of E with no limit points § This is a contradiction, so E must be bounded ○ Suppose E is not closed § Then ∃(x_0 ) ⃗∈Rk that is a limit point of E but not in E § For n∈N, ∃(x_n ) ⃗∈E s.t. |(x_n ) ⃗−(x_0 ) ⃗ | 1/n § Let S={(x_n ) ⃗ }_(n∈N □ S is infinite □ S has (x_0 ) ⃗ as a limit point § Let y ⃗∈Rk and y ⃗≠(x_0 ) ⃗ □ By triangle inequality □ |(x_n ) ⃗−y ⃗ |≥|(x_0 ) ⃗−y ⃗ |−|(x_n ) ⃗−(x_0 ) ⃗ | □ |(x_n ) ⃗−y ⃗ | |(x_0 ) ⃗−y ⃗ |−1/n □ |(x_n ) ⃗−y ⃗ | 1/2 |(x_0 ) ⃗−y ⃗ | □ For all but finitely many n □ Therefore y ⃗ cannot be a limit point of S, by Theorem 2.20 § Since y ⃗ was arbitrary, nothing other than (x_0 ) ⃗ is a limit point of S § By (c), (x_0 ) ⃗∈E,which makes a contradiction, so E has to be closed ○ Therefore E is closed and bounded Theorem 2.42 (The Weierstrass Theorem) • Statement ○ Every bounded infinite subset E of Rk has a limit point in Rk • Proof ○ E is bounded, so E⊂I⊂Rk for some k-cell I ○ By Theorem 2.40, I is compact ○ By Theorem 2.37, E has a limit point in I ○ Hence, E has a limit point in Rk Subsequences • Definition ○ Given a sequence {p_n } ○ Consider a sequence {n_k }⊂N with n_1 n_2 n_3 … ○ Then the sequence {p_(n_i ) } is a subsequence of {p_n } ○ If {p_(n_i ) } converges, its limit is called a subsequential limit of {p_n } • Example ○ Let {p_n }=1/n={1, 1/2,1/3,1/4,1/5,…} ○ One subsequence is{1, 1/4,1/6,1/7,1/38,1/101,1/135,…} ○ But{1/19,1/18,1/2,1/237,1/12,1/59,1/32,…} is not a subsequence • Note ○ A subsequential limit might exist for a sequence in the absence of a limit ○ {p_n } converges to p if and only if every subsequence of {p_n } converges to p Theorem 3.6 • Statement (a) ○ If {p_n } is a sequence in a compact metric space X ○ Then some subsequence of {p_n } converges to a point of X • Proof (a) ○ Let E be the range of {p_n } ○ If E is finite § ∃p∈E and a sequence {n_i }⊂N with n_1 n_2 n_3 … s.t. § p_(n_1 )=p_(n_2 )=p_(n_3 )=…=p ○ If E is infinite § By Theorem 2.37, E has a limit point p∈X § By Theorem 2.20, inductively choose n_i s.t. d(p,p_(n_i ) ) 1/i, ∀i∈N § It follows that {p_(n_i ) } converges to p • Statement (b) ○ Every bounded sequences in Rk contains a convergent subsequence • Proof (b) ○ By Theorem 2.41, every bounded subset of Rk is in a compact subset of Rk ○ Result follows by (a)$

Math 521 – 3/14

$Theorem 2.38 (Nested Intervals Theorem) • Statement ○ If {I_n } is a sequence of closed intervals in R s.t. I_n⊃I_(n+1),∀n∈N ○ Then ⋂24_(n=1)^∞▒I_n is nonempty • Intuition • Proof ○ Let I_n≔[a_n,b_n ] ○ Let E≔{a_n }_(n∈N § E is nonempty § E is bounded above by b_1 since b_1≥a_n,∀n∈N § So sup⁡E exists ○ Let x≔sup⁡E ○ For m,n∈N, a_n≤a_(m+n)≤b_(m+n)≤b_m § a_n≤b_m⇒x≤b_n,∀m∈N § x=sup⁡E⇒a_m≤x,∀m∈N ○ So, x∈[a_m,b_m ],∀m∈N ○ Therefore x∈⋂24_(n=1)^∞▒I_n Theorem 2.39 • Statement ○ Let k be a positive integer ○ If {I_n } is a sequence of k-cells s.t. I_n⊃I_(n+1),∀n∈N ○ Then ⋂24_(n=1)^∞▒I_n is nonempty • Proof ○ Let I_n consists of all points x ⃗=(x_1,x_2,…,x_k ) s.t. ○ a_(n,j)≤x_j≤b_(n,j), where 1≤j≤k,n=1,2,3,… ○ Let I_(n,j)=[a_(n,j),b_(n,j) ] ○ For each j, {I_(n,j) } satisfies the hypothesis of Theorem 2.38 ○ Therefore ∃x_j^∗∈⋂24_(n=1)^∞▒I_(n,j) , for 1≤j≤k ○ Let (x^∗ ) ⃗=(x_1^∗,x_2^∗,…,x_k^∗ ) ○ By construction, (x^∗ ) ⃗∈⋂24_(n=1)^∞▒I_n Theorem 2.40 • Statement ○ Every k-cell is compact • Proof ○ Let I={(x_1,x_2,…,x_k )∈Rk│a_j≤x_j≤b_j,1≤j≤k} be a k-cell ○ Let δ=√(∑_(j=1)^k▒(b_j−a_j )^2 ), then |x ⃗−y ⃗ |≤δ,∀x ⃗,y ⃗∈I ○ Suppose {G_α } is an open cover of I with no finite subcover ○ Build sequence {I_n } § Let c_j=(a_j+b_j)/2 § Consider intervals [a_j,c_j ] and [c_j,b_j ] § Those intervals describes 2^k k-cells Q_i whose union is I § Since the number of Q_i is finite, and {G_α } has no finite subcover § ∃Q_i not covered by a finite subcover of {G_α }; call this I_1 § Repeat this process on I_1 to obtain I_2,I_3,… § We can build a sequence {I_n } ○ {I_n } is a sequence of k-cells s.t. § I⊃I_1⊃I_2⊃… § I_n is not covered by any finite sub-collection of {G_α } § If x ⃗,y ⃗∈I_n, then |x ⃗−y ⃗ |≤δ/2^n ○ By Theorem 2.38, ∃x ⃗^∗∈I_n,∀n∈N ○ Then (x^∗ ) ⃗∈G_α, for some G_α § G_α is open § i.e. ∃r 0 s.t. |y ⃗−(x^∗ ) ⃗ | r⇒y ⃗∈G_α § By Archimedean Property, ∃n∈N s.t. δ/2^n r § In this case, I_n⊂G_α, which is impossible, since § I_n is not covered by any finite sub-collection of {G_α } § So no such open cover {G_α } exists ○ So every open cover of I have a finite subcover ○ Therefore I is compact$

Shawn Zhong

Shawn Zhong

Math 521

Math 521 – 3/23

Math 521 – 3/21

Math 521 – 3/19

Math 521 – 3/16

Math 521 – 3/14

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