March 5, 2018 - Shawn Zhong

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 541 - 3/5

$Proposition 24: Quotient Group • Statement ○ If G is a group, and N⊴G, then ○ the set of left costs of N, denoted as G\/N (say "G mod N") ○ is a group under the operation (g_1 N)(g_2 N)=g_1 g_2 N ○ We call this group quotient group or factor group • Proof ○ Check G\/N×G\/N→G\/N, where (g_1 N,g_2 N)↦g_1 g_2 N is well-defined § Suppose g_1 N=g_1^′ N, and g_2 N=g_2^′ N § We must show g_1 g_2 N=g_1^′ g_2^′ N § g_1 N=g_1^′ N⟺(g_1′)^(−1) g_1∈N § g_2 N=g_2^′ N⟺(g_2′)^(−1) g_2∈N § We must show that (g_1^′ g_2^′ )^(−1) g_1 g_2∈N⟺g_1 g_2 N=g_1^′ g_2^′ N § (g_1^′ g_2^′ )^(−1) g_1 g_2 § =(g_2^′ )^(−1) (g_1^′ )^(−1) g_1 g_2 § =(g_2^′ )^(−1) (g_1^′ )^(−1) g_1 [g_2^′ (g_2^′ )^(−1) ] g_2 § =(g_2^′ )^(−1) ⏟((g_1^′ )^(−1) g_1 )┬(∈N) g_2^′ ⏟((g_2^′ )^(−1) g_2 )┬(∈N) § =⏟((g_2^′ )^(−1) (g_1^′ )^(−1) g_1 g_2^′ )┬(∈N) ⏟((g_2^′ )^(−1) g_2 )┬(∈N) § Thus (g_1^′ g_2^′ )^(−1) g_1 g_2∈N § Therefore, the operation is well-defined ○ Existence of Identity § 1⋅N=N ○ Existence of Inverse § (gN)^(−1)=g^(−1) N § Since (gN)(g^(−1) N)=gg^(−1) N=N ○ Associativity § (g_1 Ng_2 N)(g_3 N) § =(g_1 g_2 N)(g_3 N) § =g_1 g_2 g_3 N § =g_1 N(g_2 g_3 N) § =g_1 N(g_2 Ng_3 N) • Note ○ If N⊴G, then there is a surjective homomorphism ○ f:G→G\/N given by g→gN ○ ker⁡f=N, since gN=N⟺g∈N ○ This shows that, if H≤G, then H⊴G iff ○ H is the kernel of a homomorphism from G to some other group • Example 1 ○ Let H be a subgroup of Z ○ H⊴Z since Z is abelian ○ Since Z is cyclic, H is also cyclic ○ So write H=⟨n⟩ ○ Then there is isomorphism § Z\/⟨n⟩→Z\/nZ § a+⟨n⟩→a ̅ • Example 2 ○ If G is a group, then {1_G }⊴G and G⊴G § G\/{1_G }≅G § G\/G≅ ∗ , where ∗ is the trivial group of order 1 ○ Intuition: The bigger the subgroup, the smaller the quotient Index • Definition ○ If G is a group, and H≤G, then ○ The index of H is the number of distinct left cosets of H in G ○ Denote the index by [G:H] • Note ○ If N⊴G, then [G:N]=|G/N| • Example ○ [Z⟨n⟩]=|ZnZ=n Theorem 25: Lagranges Theorem • Statement ○ If G is finite group, and H≤G, then |G|=|H|⋅[G:H] ○ In particular, ├ |H|┤|├ |G|┤ • Notice ○ If in the setting of Lagranges Theorem, H⊴G, then ○ |G|=|H|⋅|G/H|⇒|G/H|=|G|/|H| • Proof ○ Let n≔|H|, and k≔[G:H] ○ Let g_1,…,g_k be the representatives of the distinct cosets of H in G ○ (In other words: if g∈G, then gH∈{g_1 H,g_2 H,…,g_k H}) ○ By proposition 22, left costs are either equal or disjoint ○ So, G=g_1 H∪g_2 H∪…∪g_k H ○ Let g∈G, then there is a function f:H→gH, defined by h↦gh ○ f is certainly surjective ○ f is also injective since if gh1=gh2, then h1=h2 ○ Thus, |gH|=|H| ○ |G|=|g_1 H|+…+|g_k H|=⏟(n+n+…+n)┬(k copies)=kn ○ Therefore |G|=|H|⋅[G:H]$

Shawn Zhong

Shawn Zhong

Math 521 - 3/5

Math 521 - 3/2

Math 541 - 3/5

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