Math 521 Archives - Page 5 of 8 - Shawn Zhong

Math 521 - 3/12

$Theorem 2.34 • Statement ○ Compact subsets of metric spaces are closed • Proof ○ Let K be a compact subset of a metric space X ○ We shall prove that the complement of K is open ○ Let p∈K^c, q∈K ○ Let V_q=N_r (p),W_q=N_s (q) where r,s 1/2 d(p,q) ○ Since K is compact, ∃q_1,q_2,…q_n∈K s.t. ○ K⊂W_(q_1 )∪W_(q_2 )∪…∪W_(q_n )=W ○ Let V=V_(q_1 )∩V_(q_2 )∩…∩V_(q_n ) ○ Then V is a neighborhood of p that does not intersect W ○ V⊂K^c⇒p is an interior point of K^c ○ So K^c is open and therefore K is closed Theorem 2.35 • Statement ○ Closed subsets of compact sets are compact • Proof ○ Let X be a metric space ○ Suppose F⊂K⊂X, where F is closed, and K is compact ○ Let {V_α } be an open cover of F ○ Consider {V_α }∪{F^c }, where F^c is open ○ Then {V_α }∪{F^c } is an open cover of K ○ Since K is compact, K has a finite subcover Φ ○ If F^c∈Φ, then Φ∖{F^c } is still finite and covers F ○ So we have a finite subcover of {V_α } ○ Therefore F is compact • Corollary ○ If F is closed and K is compact , then F∩K is compact • Proof ○ K compact ⇒K is closed ○ We know F is closed, so F∩K is closed ○ F∩K⊂K, and K is compact ○ So F∩K is compact Theorem 2.36 (Cantor s Intersection Theorem) • Statement ○ If {K_α } is a collection of compact subsets of a metric space X s.t. ○ The intersection of every finite subcollection of {K_α } is nonempty ○ Then ⋂_α▒K_α is nonempty • Proof ○ Fix K_1∈{K_α } and let G_α=K_α^c \,∀α ○ Assume no point of K_1 belongs to every K_α ○ Then {G_α } is an open cover of K_1 ○ Since K_1 is compact, K_1⊂G_(α_1 )∩G_(α_2 )∩…∩G_(α_n ) ○ Where α_1,α_2,…,α_n is a finite collection of indices ○ Then K_1∩G_(α_2 )∩…∩G_(α_n )=∅ ○ This is a contradiction, so no such set K_1 exists ○ The result follows • Corollary ○ If {K_n } is a sequence of nonempty compact sets s.t. K_n⊃K_(n+1),∀n∈N ○ Then ⋂24_(n=1)^∞▒K_n is nonempty Theorem 2.37 • Statement ○ If E is an infinite subset of a compact set K ○ Then E has a limit point in K • Proof ○ If no point of K were a limit point of E ○ Then ∀q∈K, ∃N(q) s.t. no point of E other than q ○ i.e. N(q) contains at most one point of E (namely, q, if q∈E) ○ So no finite sub-collection of {N(q)} can cover E, and thus not K ○ This is a contradiction, so E has a limit point in K$

Math 521 - 3/7

$Theorem 3.4 • Statement (a) ○ Suppose (x_n ) ⃗=(α_(1,n),α_(2,n),…,α_(k,n) )∈Rk where n∈N, then ○ {(x_n ) ⃗ } converges to (α_1,α_2,…,α_k )⟺(lim)_(n→∞)⁡〖α_(j,n) 〗=α_j (1≤j≤k) • Proof (a) ○ Assume (x_n ) ⃗→x ⃗ § Given ε 0, there exists N∈N s.t. |(x_n ) ⃗−x ⃗ | ε for n≥N § Thus, |α_(j,n)−α_j |≤|(x_n ) ⃗−x ⃗ | for n≥N,1≤j≤k § Therefore lim_(n→∞)⁡〖α_(j,n) 〗=α_j for 1≤j≤k ○ Assume lim_(n→∞)⁡〖α_(j,n) 〗=α_j for 1≤j≤k § Given ε 0, there exists N∈N s.t. |α_(j,n)−α_j | ε/√k for n≥N § |(x_n ) ⃗−x ⃗ |=|√(∑_(i=1)^k▒|α_(j,n)−α_n |^2 )|=√(∑_(i=1)^k▒|α_(j,n)−α_n |^2 ) √(∑_(i=1)^k▒ε^2/k)=ε § Therefore (x_n ) ⃗→x ⃗ • Statement (b) ○ Suppose § {(x_n ) ⃗ } and {(y_n ) ⃗ } are sequences in Rk, {β_n } is a sequence in R § (x_n ) ⃗→x ⃗, (y_n ) ⃗→y ⃗, β_n→β ○ Then § (lim)_(n→∞)⁡〖(x_n ) ⃗+(y_n ) ⃗ 〗=x ⃗+y ⃗ § (lim)_(n→∞)⁡〖(x_n ) ⃗⋅(y_n ) ⃗ 〗=x ⃗⋅y ⃗ § (lim)_(n→∞)⁡〖β_n⋅(x_n ) ⃗ 〗=β⋅x ⃗ • Proof (b) ○ This follows from (a) and Theorem 3.3 (Algebraic Limit Theorem) Compact Sets • Intuition for Rk: Closed and bounded • Open cover ○ An open cover of a set E in a metric X is ○ a collection of open sets {G_α } in X s.t. E⊂⋃8_α▒G_α • Compact ○ A set K in a metric space X is compact if ○ every open cover of K has a finite subcover • Example 1 ○ Let E=(0,1), X=R ○ E is a open cover of itself, but E is not compact ○ Let G_α=(α/2,1) for α∈(0,1), then E has {G_n } as an open cover ○ We cannot take a finite collection of these G_α and still have an open cover ○ So it has no finite subcover ○ Therefore E=(0,1) is not compact • Example 2 ○ Let K=[0,1], X=R ○ Consider {G_α }∪{G_0 }∪{G_1 }, where § G_α=(α/2,1) for α∈(0,1) § G_0=(−ε,ε) § G_1=(1−ε,1+ε) for some ε 0 ○ Then {G_α }∪{G_0 }∪{G_1 } is an open cover of [0,1] ○ It has finite subcover {G_0,G_1,G_ε } where G_ε=(ε/2,1) ○ Therefore K=[0,1] is compact$

Math 521 - 3/5

$Theorem 3.3 (Algebraic Limit Theorem) • Suppose {s_n },{t_n } are complex sequence, and lim_(n→∞)⁡〖s_n 〗=s,lim_(n→∞)⁡〖t_n 〗=t, then • (lim)_(n→∞)⁡〖s_n+t_n 〗=s+t ○ Given ε 0 § lim_(n→∞)⁡〖s_n 〗=s⇒∃N_1∈N s.t. |s_n−s| ε/2 for n≥N_1 § lim_(n→∞)⁡〖t_n 〗=t⇒∃N_2∈N s.t. |t_n−t| ε/2 for n≥N_2 ○ Let N=max⁡(N_1,N_2 ), then for n≥N § |s_n+t_n−(s+t)|=|(s_n−s)+(t_n−t)|≤|s_n−s|+|t_n−t| ε ○ Therefore lim_(n→∞)⁡〖s_n+t_n 〗=s+t • (lim)_(n→∞)⁡〖c+s_n 〗=c+s, ∀c∈ℂ ○ Given ε 0 ○ lim_(n→∞)⁡〖s_n 〗=s⇒∃N∈N s.t.|s_n−s| ε for n≥N ○ So, |c+s_n−(c+s)|=|s_n−s| ε ○ Therefore lim_(n→∞)⁡〖c+s_n 〗=c+s • (lim)_(n→∞)⁡〖cs_n 〗=cs, ∀c∈ℂ ○ Given ε 0 ○ If c=0 § |cs_n−cs|=0 ε ○ If c≠0 § lim_(n→∞)⁡〖s_n 〗=s⇒∃N∈N s.t. |s_n−s| ε/|c| for n≥N § So |cs_n−cs|=|c||s_n−s| |c| ε/|c| =ε ○ Therefore lim_(n→∞)⁡〖cs_n 〗=cs • (lim)_(n→∞)⁡〖s_n t_n 〗=st ○ Standard approach § s_n t_n−st=s_n t_n−st_n+st_n−st=t_n (s_n−s)+s(t_n−t) ○ Rudin s approach § s_n t_n−st=(s_n−s)(t_n−t)+t(s_n−s)+s(t_n−t) ○ Given ε 0 § ∃N_1∈N s.t. |s_n−s| √ε for n≥N_1 § ∃N_2∈N s.t. |t_n−t| √ε for n≥N_2 ○ Let N=max⁡(N_1,N_2 ),then § |(s_n−s)(t_n−t)| ε for n≥N § ⇒lim_(n→∞)⁡(s_n−s)(t_n−t)=0 ○ lim_(n→∞)⁡〖s_n t_n 〗 § =lim_(n→∞)⁡[(s_n−s)(t_n−t)+t(s_n−s)+s(t_n−t)+st] § =lim_(n→∞)⁡(s_n−s)(t_n−t)+t lim_(n→∞)⁡(s_n−s)+s lim_(n→∞)⁡(t_n−t)+st § =0+0+0+st § =st ○ Therefore lim_(n→∞)⁡〖s_n t_n 〗=st • (lim)_(n→∞)⁡〖1/s_n 〗=1/s (s_n≠0, ∀n∈N and s≠0) ○ lim_(n→∞)⁡〖s_n 〗=s⇒∃N^′∈N s.t. |s_n−s| |s|/2 for n≥N^′ ○ By the Triangle Inequality, |s|−|s_n |≤|s_n−s|,∀n≥N^′ ○ ⇒|s_n |≥|s|−|s_n−s| |s|−|s|/2=|s|/2,∀n≥N^′ ○ Given ε 0, ∃N N^′ s.t. |s_n−s| 1/2 |s|^2 ε for n≥N ○ |1/s_n −1/s|=|(s−s_n)/(s_n s)| (1/2 |s|^2 ε)/(|s_n |⋅|s| ) (1/2 |s|^2 ε)/(|s|/2⋅|s| )=ε ○ Therefore lim_(n→∞)⁡〖1/s_n 〗=1/s$

Math 521 - 3/2

$Theorem 3.2 • Let {p_n } be a sequence in a metric space X • p_n→p∈X⟺ any neighborhood of p contains p_n for all but finitely many n ○ Suppose {p_n } converges to p § Let B be a neighborhood of p with radius ε § p_n→p⇒∃N∈N s.t.d(p_n,p) ε,∀n≥N § So, p_n∈B,∀n≥N § p_1,…,p_(n−1) may not be in B, but there are only finitely many of these ○ Suppose every neighborhood of p contains all but finitely many p_n § Let ε 0 be given § B≔{q∈X│d(p,q) ε} is a neighborhood of p § By assumption, all but finitely points in {p_n } are in B § Choose N∈N s.t. N i,∀p_i∉B § Then d(p_n,p) ε,∀n≥N § So, lim_(n→∞)⁡〖p_n 〗=p • Given p∈X and p^′∈X. If {p_n } converges to p and to p′, then p=p^′ ○ Let ε 0 be given § {p_n } converges to p⇒∃N∈N s.t. d(p_n,p) ε/2,∀n≥N_1 § {p_n } converges to p^′⇒∃N^′∈N s.t. d(p_n,p^′ ) ε/2,∀n≥N_2 ○ Let N=max⁡(N_1,n_2 ), then § d(p,p^′ )≤d(p_n,p)+d(p_n,p^′ ) ε/2+ε/2=ε,∀n≥N ○ Since ε 0 is arbitrary, d(p,p^′ )=0 ○ Therefore p=p^′ • If {p_n } converges, then {p_n } is bounded ○ Since {p_n } converges to some p ○ Let ε=1, then ∃N∈N s.t. d(p_n,p) 1 ○ Let q=max⁡(1,d(p_1,p),d(p_2,p),…,d(p_(N−1),p)) ○ Then d(p,p_n ) q,∀n∈N ○ By definition, {p_n } is bounded • If E⊂X, and p∈E^′, then there exists a sequence {p_n } in E s.t.p_n→p ○ Since p is a limit point of E ○ Every neighborhood of p contains q≠p, and q∈E ○ Consequently, ∀n∈N, ∃p_n∈E s.t. d(p_n,p) 1/n ○ Let ε 0 be given ○ By Archimedean property, ∃N∈N s.t. 1/N ε ○ So for n≥N, 1/n ε⇒d(p_n,p) 1/n ε ○ Therefore p_n→p$

Math 521 - 2/28

$Convergence and Divergence • Definition ○ A sequence {p_n } in a metric space X converges to a point p∈X if ○ Given any ε 0, ∃N∈N s.t. d(p,p_n ) ε, ∀n≥N ○ If {p_n } converges to p, we write § p_n→p § lim_(n→∞)⁡〖p_n 〗=p § lim⁡〖p_n 〗=p ○ If {p_n } does not converge, it is said to diverge • Intuition ○ ε is small ○ N is a "point of no return" beyond which sequence is within ε of p Range and Bounded • Range ○ Given a sequence {p_n } ○ The set of points p_n (n∈N) is called the range of the sequence ○ Range could be infinite, but it is always at most countable ○ Since we can always construct a function f:N→{p_n }, where f(n)=p_n • Bounded ○ A sequence {p_n } is said to be bounded if its range is bounded Examples • Consider the following sequences of complex numbers {s_n } Limit Range Bounded s_n=1/n 0 Infinite Yes s_n=n^2 Divergent Infinite No s_n=1+(−1)^n/n 1 Infinite Yes s_n=i^n Divergent {±1,±i} Yes s_n=1 1 {1} Yes • Proof:lim_(n→∞)⁡〖1/n〗=0 ○ Let ε 0 ○ By Archimedean Property, we can choose N∈N s.t. N 1/ε ○ ∀n≥N, n 1/ε⇒1/n ε ○ i.e. d(1/n,0)=|1/n| ε,∀n≥N ○ Therefore lim_(n→∞)⁡〖1/n〗=0$

Shawn Zhong

Shawn Zhong

Math 521

Math 521 - 3/12

Math 521 - 3/7

Math 521 - 3/5

Math 521 - 3/2

Math 521 - 2/28

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