Notes Archives - Page 17 of 84 - Shawn Zhong

Math 521 – 4/20

$Uniform Continuity • Let X,Y be metric spaces, f:X→Y • f is uniformly continuous on X if for every ε 0, ∃δ 0 s.t. • If p,q∈X and d_X (p,q) δ, then d_Y (f(p),f(q)) ε Theorem 4.19 • Statement ○ Let X,Y be metric spaces, X compact ○ If f:X→Y is continuous, then f is also uniformly continuous • Proof ○ Let ε 0 be given ○ Since f is continuous, ∀p∈X, ∃ϕ(p) s.t. § If q∈X, and d_X (p,q) ϕ(p), then d_Y (f(p),f(q)) ε/2 ○ Let J(p)≔{q∈X│d_X (p,q) 1/2 ϕ(p) } § p∈J(p),∀p∈X, so {J(p)} is an open cover of X § Since X is compact, {J(p)} has a finite subcover § There exists finite set of points p_1,…,p_n∈X s.t. § X⊂J(p_1 )∪…∪J(p_n ) ○ Let δ=1/2 min⁡{ϕ(p_1 ),…,ϕ(p_n )} 0 § Given p,q∈X s.t. d_X (p,q) δ § Since X⊂J(p_1 )∪…∪J(p_n ), § ∃m∈{1,2,…,n} s.t. p∈J(p_m ) ○ Hence § d_X (p,p_m ) 1/2 ϕ(p_m ) ϕ(p_m ) § d_X (q,p_m )≤d_X (p,q)+d_X (p,p_m ) δ+1/2 ϕ(p_m )≤ϕ(p_m ) ○ By the triangle inequality and definition of ϕ(p), ○ d_Y (f(p),f(q))≤d_Y (f(p),f(p_m ))+d_Y (f(p_m ),f(q)) ε/2+ε/2=ε Theorem 4.20 • Definition ○ Let E be noncompact set in R ○ Then there exists a continuous function f on E s.t. (a) f is not bounded (b) f is bounded but has no maximum (c) E is bounded, but f is not uniformly continuous • Proof : If E is bounded ○ Since E is noncompact, E must be not closed ○ So there exists a limit point x_0∈E s.t. x_0∉E ○ f(x)≔1/(x−x_0 ) establishes (c) § f is continuous by Theorem 4.9 § f is clearly unbounded § f is not uniformly continuous □ Let ε 0 and δ 0 be arbitrary □ Choose x∈E s.t. |x−x_0 | δ □ Taking t close to x_0 □ We can make |f(t)−f(x)| ε, but |t−x| δ □ Since δ 0 is arbitrary ○ g(x)≔1/(1+(x−x_0 )^2 ) establishes (b) § g is continuous by Theorem 4.9 § g is bounded, since 0 g(x) 1 § g has no maximum, since sup┬(x∈E)⁡g(x)=1, but g(x) 1 • Proof: If E is not bounded ○ f(x)≔x establishes (a) ○ h(x)≔x^2/(1+x^2 ) establishes (b) Example 4.21 • Let X=[0,2π) • Let f:X→Y given by f(t)=(cos⁡t,sin⁡t ) • Then f is continuous, and bijective • But f^(−1) is not continuous at f(0)=(1,0)$

Math 521 – 4/18

$Bounded • A mapping f:E→Rk is bounded if • there is a real number M s.t. |f(x)|≤M,∀x∈E Theorem 4.14 • Statement ○ Let X,Y be metric spaces, X compact ○ If f:X→Y is continuous, then f(X) also is compact • Proof ○ Let {V_α } be an open cover of f(X) ○ f is continuous, so each of the sets f^(−1) (V_α ) is open by Theorem 4.8 ○ {f^(−1) (V_α )} is an open cover of X, and X is compact ○ So there is a finite set of indices {α_1,α_2,…,α_n } s.t. ○ X⊂f^(−1) (V_(α_1 ) )∪f^(−1) (V_(α_2 ) )∪…∪f^(−1) (V_(α_n ) ) ○ Since f(f^(−1) (E))⊂E,∀E⊂Y ○ f(X)⊂V_(α_1 )∪V_(α_2 )∪…∪V_(α_n ) ○ This is a finite subcover of f^(−1) (X) Theorem 4.15 • Statement ○ Let X be a compact metric space ○ If f:X→Rk is continuous, then f(X) is closed and bounded ○ Thus, f is bounded • Proof ○ See Theorem 4.14 and Theorem 2.41 Theorem 4.16 (Extreme Value Theorem) • Statement ○ Let f be a continuous real function on a compact metric space X ○ Let M≔sup┬(p∈X)⁡f(p), m≔inf┬(p∈X)⁡f(p) ○ Then ∃p,q∈X s.t. f(p)=M and f(q)=m ○ Equivalently, ∃p,q∈X s.t. f(q)≤f(x)≤f(p),∀x∈X • Proof ○ By Theorem 4.15, f(X) is closed and bounded ○ So f(x) contains M and m by Theorem 2.28 Theorem 4.17 • Statement ○ Let X,Y be metric spaces, X compact ○ Suppose f:X→Y is continuous and bijictive ○ Define f^(−1):Y→X by f^(−1) (f(x))=x,∀x∈X ○ Then f^(−1) is also continuous and bijective • Proof ○ By Theorem 4.8 applied to f^(−1) ○ It suffices to show f(V) is open in Y for all open sets V in X ○ Fix an open set V in X ○ V is open in compact metric space X ○ So V^c is closed and compact by Theorem 2.35 ○ Therefore, f(V^c ) is a compact subset of Y by Theorem 4.14 ○ So f(V^c ) is closed in Y by Theorem 2.34 ○ f is 1-1 and onto, so f(V)=(f(V^c ))^c ○ Therefore f(V) is open$

Math 521 – 4/16

$Continuous • Definition ○ Suppose X,Y are metric spaces, E⊂X, p∈E, and f:E→Y ○ f is continuous at p if for every ε 0, there exists δ 0 s.t. ○ If x∈E,d_X (x,p) δ, then d_Y (f(x),f(p)) ε ○ If f is continuous at every point p∈E, then f is continuous on E • Note ○ f must be defined at p to be continous at p (as opposed to limit) ○ If p is an isolated point of E ○ Then every function f whose domain is E is continous at p Theorem 4.6 • In the context of Definition 4.5, if p is also a limit point of E, then • f is continious at p if and only if (lim)_(x→p)⁡f(x)=f(p) Theorem 4.7 • Statement ○ Suppose X,Y,Z are metric spaces, E⊂X,f:E→Y, g:f(E)→Z, and ○ h:E→Z defined by h(x)=g(f(x)),∀x∈E ○ If f is continuous at p∈E, and g is continuous at f(p) ○ Then h is continuous at p • Note: h is called the composition of f and g and is called g∘f • Proof ○ Let ε 0 be given ○ Since g is continuous at f(p),∃η 0 s.t. § If y∈f(E) and d_Y (y,f(p)) η, then d_Z (g(y),g(f(p))) ε ○ Since f is continuous at p, ∃δ 0 s.t. § If x∈E and d_X (x,p) δ, then d_Y (f(x),f(p)) η ○ Consequently, if d_X (x,p) δ, and x∈E, then § d_Z (g(f(x)),g(f(p)))=d_Z (hx),hp)) ε ○ So, h is continuous at p by definition Theorem 4.8 • Statement ○ Given metric spaces X,Y ○ f:X→Y is continuous if and only if ○ f^(−1) (V) is open in X for every open set V⊂Y • Proof (⟹) ○ Suppose f is continuous on X, and V⊂Y is open ○ We want to show all points of f^(−1) (V) are interior points ○ Suppose p∈X, and f(p)∈V, then p∈f^(−1) (V) ○ V is open, so ∃ε 0 s.t. y∈V if d_Y (f(p),y) ε ○ Since f is continuous at p, ∃δ 0 s.t. § If d_X (x,p) δ, then d_Y (f(x),f(p)) ε ○ So x∈f^(−1) (V) if d_X (x,p) δ ○ This shows that p is an interior point of f^(−1) (V) ○ Therefore f^(−1) (V) is open in X • Proof (⟸) ○ Suppose f^(−1) (V) is open in X for every open set V⊂Y ○ Fix p∈X, ε 0 ○ Let V≔{y∈Y│d_Y (y,f(p)) ε } ○ V is open, so f^(−1) (V) is also open ○ Thus, ∃δ 0 s.t. if d_X (p,x) δ, then x∈f^(−1) (V) ○ But if x∈f^(−1) (V), then f(x)∈V so d_Y (f(x),f(p)) ε ○ So, f:X→Y is continuous at p ○ Since p∈X was arbitrary, f is continuous on X • Corollary ○ Given metric spaces X,Y ○ f:X→Y is continuous on X if and only if ○ f^(−1) (V) is closed in X for every closed set V in Y • Proof ○ A set is closed if and only if its complement is open ○ Also, f^(−1) (E^c )=[f^(−1) (E)]^c, for every E⊂Y$

Math 521 – 4/11

$Theorem 3.55 • Statement ○ If Σa_n is a series of complex numbers which converges absolutely ○ Then every rearrangement of Σa_n converges to the same sum • Proof ○ Let Σa_n^′ be a rearrangement of Σa_n with partial sum s_n^′ ○ By the Cauchy Criterion, given ε0, ∃N∈N s.t. § ∑_(i=n)^m▒|a_i | ε,∀m,n≥N ○ Choose p s.t. 1,2,…,N are all contained in the set {k_1,k_2,…,k_p } ○ Where k_1,…,k_p are the indices of the rearranged series ○ Then if np, a_1,…,a_N will be cancelled in the difference s_n−s_n^′ ○ So, |s_n−s_n^′ |≤ε⇒{s_n^′ } converges to the same value as {s_n } Limit of Functions • Definition ○ Let X,Y be metric spaces, and E⊂X ○ Suppose f:E→Y and p is a limit point of E ○ We write § f(x)→q as x→p, or § (lim)_(x→p)⁡f(x)=q ○ If ∃q∈Y s.t. § Given ε0, there exists δ0 s.t. § If 0d_X (x,p)δ, then d_Y (f(x),q)ε • Note ○ 0d_X (x,p)δ is the deleted neighborhood about p of radius δ ○ d_X and d_Y refer to the distances in X and Y, respectively • Relationship with sequence ○ Theorem 4.2 relates this type of limit to the limit of a sequence ○ Consequently, if f has a limit at p, then its limit is unique Theorem 4.3 • If f,g are complex function on E, we have • (f+g)(x)=f(x)+g(x) • (f−g)(x)=f(x)−g(x) • (fg)(x)=f(x)g(x) • (f/g)(x)=f(x)/g(x) where g(x)≠0 on E Theorem 4.4 (Algebraic Limit Theorem) • Let X be a metric space, E⊂X • Suppose p be a limit point of E • Let f,g be complex functions on E where ○ lim_(x→p)⁡f(x)=A ○ lim_(x→p)⁡g(x)=B • Then ○ lim_(x→p)⁡(f(x)+g(x))=A+B ○ lim_(x→p)⁡(f(x)g(x))=AB ○ lim_(x→p)⁡(f(x)/g(x) )=A/B where B≠0$

Math 541 – 4/18

$Proposition 64 • Statement ○ Let n 0 ○ Every nonzero element in Z\/nZ is either a unit or a zero-divisor • Note ○ We don’t have this property in Z ○ In Z, the units are ±1, there are no zero-divisor ○ 2∈Z is not 0 or unit or zero-divisor • Proof ○ Suppose a ̅∈Z\/nZ is nonzero and not a unit ○ Then d≔(a,n) 1 ○ Write cd=a,md=n ○ Then a ̅m ̅=c ̅d ̅m ̅=c ̅n ̅=0 ̅ ○ Moreover, m ̅≠0 ̅ § Since md=n,1≤m≤n, and d 1 § m cannot be a multiple of n Field • Definition ○ Communitive ring R is called a field if ○ Every nonzero element of R is a unit ○ i.e. Every nonzero element of R have a multiplicative inverse • Examples ○ Q,R ○ ℂ § But not true for R2 with (r_1,r_2 )(r_1^′,r_2^′ )=(r_1 r_1^′,r_2 r_2^′ ) ○ Z\/pZ (p prime) § 1≤a≤p−1,(a,p)=1⇒a ̅∈Z\/pZ § Note: Z\/nZ is a field ⟺ n is prime Product Ring • If R_1,R_2 are rings, R_1×R_2 has the following ring structure • For addition, it s just the product as groups • For multiplication, (r_1,r_2 )(r_1^′,r_2^′ )=(r_1 r_1^′,r_2 r_2^′ ) with identity (1_(R_1 ),1_(R_2 ) ) Integral Domain • Definition ○ A communicative ring R is an integral domain (or just domain) if ○ R contains no zero-divisors • Example ○ Unites are not zero-divisors, so fields are domains ○ Z is a domain ○ Z\/nZ is a domain ⟺ it is a field ○ R_1×R_2 is a domain ⟺ one of them is trivial, and the other is a domain$

Shawn Zhong

Shawn Zhong

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Math 521 – 4/20

Math 521 – 4/18

Math 521 – 4/16

Math 521 – 4/11

Math 541 – 4/18

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