April 22, 2018 - Shawn Zhong

Math 541 – 4/20

$Product Ring and Domain • Statement ○ If R_1 and R_2 are rings, then R_1×R_2 is a domain iff ○ one of the R_1 or R_2 is a domain, and the other is trivial • Proof (⟸) ○ Without loss of generality, assume R_1 is a domain and R_2 is trivial ○ Let (r_1,r_2 ),(r_1^′,r_2^′ )∈R_1×R_2∖{(0,0)} ○ Then r_1≠0 and r_1^′≠0 ○ Since R_1 is a domain, r_1 r_1^′≠0 ○ Thus, (r_1,r_2 )(r_1^′,r_2^′ )=(r_1 r_1^′,r_2 r_2^′ )≠0 • Proof (⟹) ○ (1_(R_1 ),0)(0,1_(R_2 ) )=(1_(R_1 )⋅0,0⋅1_(R_2 ) )=(0,0) ○ Since R_1×R_2 is a domain, either (1_(R_1 ),0) or (〖0,1〗_(R_2 ) ) is (0,0) ○ This means either 1_(R_1 ) or 1_(R_2 ) is 0, and thus R_1 or R_2 is trivial ○ Without loss of generality, suppose R_2 is trivial ○ We want to show that R_1 is a domain ○ Let r_1,r_1^′∈R_1∖{0} ○ Then (r_1,0),(r_1^′,0)∈R_1×R_2∖{(0,0)} ○ So (r_1 r_1^′,0)≠0 i.e. r_1 r_1^′≠0 Proposition 66 • Statement ○ A finite domain R is a field • Proof ○ Let a∈R∖{0} ○ We want to show that a has a multiplicative inverse ○ Define a function F:R→R given by r↦ar ○ F is injective § If ar_1=ar_2 § Then a(r_1−r_2 )=0 § Since R is a domain, r_1−r_2=0 § So r_1=r_2 ○ F is surjective since R is finite ○ Choose b∈R s.t. F(b)=1, then ab=1 ○ So b is the inverse of a Subring • Definition ○ A subring of a ring R is a additive subgroup S of R s.t. ○ S is closed under multiplication ○ S contains 1 • Note: A subring of a ring is also a ring • Example 1 ○ A ring is always a subring of itself • Example 2 ○ Mat_n (R has a subring given by diagonal matrices ○ Scalar matices also form a subring • Example 3 ○ Z⊆Q⊆R⊆ℂ is a chain of subring of ℂ • Example 4 ○ R={continuous functions from Rn to Rfor some n≥1} ○ Addition: (f+g)(v)=f(v)+g(v) ○ Multiplication: (fg)(v)=f(v)g(v) with identity of constant function 1 ○ Polynomials in n variables form a subring • Example 5 ○ If f:R→S is a ring homomorphism i.e. § f is a homomorphism of abelian groups under addition § f(r_1 r_2 )=f(r_1 )f(r_2 ),∀r_1,r_2∈R § f(1_R )=1_S ○ Then im(f) is a subring of S ○ Proof § By group theory, im(f) is an additive subgroup of S § 1∈im(f) by assumption § If f(r_1 ),f(r_2 )∈im(f), then f(r_1 )f(r_2 )=f(r_1 r_2 )∈im(f) • Example 6 ○ By HW, ∃! Ring homomorphism f:Z→R for any ring R ○ im(f) is the smallest subring of R ○ Also, im(f)≅Z\/nZ, where n=char(R) ○ Note: A ring isomorphism is a ring homomorphism that is bijective • Example 7 ○ {(r_1,0)│r_1∈R_1 }⊆R_1×R_2 is not a subring ○ Since it doesn t contain the identity (1,1)$

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$

Shawn Zhong

Shawn Zhong

Math 541 – 4/20

Math 521 – 4/20

Math 521 – 4/18

Math 521 – 4/16

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