Math 521 Archives - Page 8 of 8 - Shawn Zhong

Math 521 - 1/31

$Upper Bound and Lower Bound • Suppose S is an ordered set and E⊂S • If there exists β∈S such that x≤β, ∀x∈E • We say that x is bounded above and call β an upper bound for E • If there exists β∈S such that x≥β, ∀x∈E • We say that x is bonded below by β, and β is a lower bound for E Least Upper Bound and Greatest Lower Bound • Definition ○ Suppose S is an ordered set and E⊂S is bounded above. ○ Suppose there exists α∈S s.t. § α is an upper bond of E § If γ α, then γ is not an upper bound of E ○ Then we call α the least upper bound (or lub or sup or supremium) of E ○ Suppose there exists α∈S s.t. § α is an lower bond of E § If γ α, then γ is not an lower bound of E ○ Then we call α the greastst lower bound (or glb or inf or infimum) of E • Examples ○ Recall § A={q∈Qq^2 2} has no sup in Q § B={q∈Qq^2 2} has no inf in Q ○ If α=sup⁡E exists, α may or may not be in E § E_1≔{r∈Qr 0} □ inf⁡〖E_1 〗 doesn t exist □ sup⁡〖E_1 〗=0∉E_1 § E_2≔{r∈Qr≤0} □ inf⁡〖E_2 〗 doesn t exist □ sup⁡〖E_2 〗=0∈E_2 § E≔{1/n│n∈N={1, 1/2,1/3,1/4,⋯} □ inf⁡E=0∉E □ sup⁡E=1∈E • Least-upper-bound property ○ We say that a ordered set S has least-upper-bound property provided that ○ if E∈S s.t. E≠∅ and E is bounded above, then sup⁡E exists and sup⁡E∈S Theorem 1.11 • Statement ○ Suppose S is an ordered set with the least-upper-bound property ○ Suppose B⊂S, B≠∅ and B is bounded below ○ Let L be the set of lower bounds of B ○ Then α=sup⁡L exists in S and α=inf⁡B • Proof ○ L≠∅ § B is bounded below, so L is not empty ○ L is bounded above § Given b∈B and l∈L, we have l≤b by definition of L § Therefore, L is bounded above ○ sup⁡L exists in S § L≠∅, L is bounded above § And S has least upper bound property § So sup⁡L exists § Let α=sup⁡L∈S ○ α is a lower bound for B (i.e. α∈L) § If γ α, then γ is not an upper bound for L, so γ∉B § So α≤x for all x∈B § Thus, α is a lower bound for B § i.e. α∈L ○ α=inf⁡B § If β α is another lower bound for B § Then β∉L since α is an upper bound for L § So, α∈L, but β∉L if β α § Therefore α is the least upper bound of B § i.e. α=inf⁡B ○ Therefore α=sup⁡L=inf⁡B∈S Ordered Field • Definition ○ An ordered field is a field F which is also an ordered set, such that § x+y x+z if x,y,z∈F and y z § xy 0 if x,y∈F, x 0 and y 0 ○ If x 0, we call x positive ○ If x 0, we call x negative • Examples ○ N,Z,Q, R • Note ○ R is an ordered field with least-upper-bound property$

Math 521 - 1/29

Proposition 1.14 • Given a field F, for x,y,z∈F (1) If x+y=x+z, then y=z (2) If x+y=x, then y=0 (3) If x+y=0, then y=−x (4) −(−x)=x • Proof (1) ○ x+y=x+z ○ (x+y)+(−x)=(x+z)+(−x) by (A5) ○ x+y+(−x)=x+z+(−x) by (A3) ○ x+(−x)+y=x+(−x)+z by (A2) ○ 0+y=0+z by (A6) ○ y=z∎ by (A4) • Proof (2) ○ x+y=x=x+0 ○ ⇒y=0∎ • Proof (3) ○ x+y=0=x+(−x) ○ ⇒y=−x∎ • Proof (4) ○ (−x)+(−(−x))=0 ○ x+(−x)+(−(−x))=x+0 ○ 0+(−(−x))=x+0 ○ −(−x)=x∎ Proposition 1.15 • Given a field F, for x,y,z∈F (1) If x≠0 and xy=xz, then y=z (2) If x≠0 and xy=x, then y=1 (3) If x≠0 and xy=1, then y=1/x (4) If x≠0, then 1/(1/x)=x • Proof similar to Proposition 1.14 Proposition 1.16 • Given a field F, for x,y∈F (1) 0x=0 (2) If x≠0 and y≠0, then xy≠0 (3) (−x)y=−(xy)=x(−y) (4) (−x)(−y)=xy • Proof (1) ○ 0+0=0 ○ (0+0)x=0x ○ 0x+0x=0x ○ 0x+0x+(−(0x))=0x+(−(0x)) ○ 0x=0∎ • Proof (2) ○ Suppose x≠0, y≠0, but xy=0 ○ x≠0, so 1/x exists ○ 1/x (xy)=1/x⋅0 ○ (1/x⋅x)y=1/x⋅0 ○ 1⋅y=0 ○ y=0 ○ This is a contradiction, so xy≠0∎ • Proof (3) ○ (−x)y+xy=((−x)+x)y=0⋅y=0 ○ (−x)y+xy+(−xy)=0+(−xy) ○ (−x)y=−xy ○ And the rest is similar • Proof (4) ○ Use (3), (−x)(−y)=−(x(−y))=−(−xy)=xy∎ Order • Intuition ○ The real number line • Definition ○ Let S be a set. ○ An order on S is a relation, denoted by ,with the following two properties: § If x∈S and y∈S, then one and only one of the statements xy, x=y, yx is true § If x,y,z∈S, if xy and yz, then xz (Transitivity) ○ x≤y means either xy or x=y ○ x≥y means either xy or x=y ○ An ordered set is a set for which an order is defined. • Example ○ Q is an ordered set under the definition that ○ For r,s∈Q, rs, if and only if s−r is positive Upper Bound and Lower Bound • Definition ○ Suppose S is an ordered set and E⊂S. ○ If there exists β∈S such that x≤β, ∀x∈E ○ We say that x is bounded above and call β an upper bound for E ○ Similarly, if x≥β, ∀x∈E. ○ We say that x is bonded below by β, and β is a lower bound for E

Math 521 - 1/26

$Sets • Contains ○ If A is a set and x is an element of A (an object of A), then we write x∈A ○ Otherwise, we write x∉A • Set ○ The empty set or null set is a set with no elements, and is denoted as ∅ ○ If a set has at least one element, it is called nonempty • Subset ○ If A and B are sets and every element of A is an element of B ○ Then A is a subset of B ○ Rubin write this A⊂B, or B⊃A ○ A⊂A for all sets A • Proper subset ○ If B contain something not in A, then A is a proper subset of B • Equal ○ If A⊂B and B⊂A then A=B. ○ Otherwise A≠B √2 is Not Rational • We proved that √2 is not rational • i.e. there is no rational number p such that p^2=2 • Let A={p∈Qp^2<2}, B={p∈Qp^2>2} • Prove: A has no largest element, and B has no smallest element ○ Let p∈Q, and p>0 ○ Let q≔p−(p^2−2)/(p+2)=(2p+2)/(p+2),then q^2−2=((2p+2)/(p+2))^2−2=2(p^2−2)/(p+2)^2 ○ If p∈A § then p^2−2<0 § ⇒q^2−2=2(p^2−2)/(p+2)^2 <0 § ⇒q^2<2 § ⇒q∈A § ⇒q>p § i.e. A has no largest element ○ If p∈B § then p^2−2>0 § ⇒q^2−2=2(p^2−2)/(p+2)^2 >0 § ⇒q^2>2 § ⇒q∈B § ⇒q</p> <p § i.e. B has no smallest element What is R? • The real numbers are an example of field. • A field is a set F with two binary operations called addition and multiplication that satisfy that following axioms: • Axioms for addition (+) ○ (A1) If x∈F and y∈F, then x+y∈F ○ (A2) Addition is communicate: x+y=y+x,∀x,y∈F ○ (A3) Addition is associative: (x+y)+z=x+(y+z),∀x,y,z∈F ○ (A4) There exists 0∈F s.t. x+0=x, ∀x∈F ○ (A5) ∀x∈F, there exists an additive inverse −x∈F s.t. x+(−x)=0 • Axioms for multiplication (× or ⋅) ○ (M1) If x∈F and y∈F, then xy∈F ○ (M2) Addition is communicate: xy=yx,∀x,y∈F ○ (M3) Addition is associative: (xy)z=x(yz),∀x,y,z∈F ○ (M4) F contains an element 1≠0 s.t. 1⋅x=x, ∀x∈F ○ (M5) If x∈F and x≠0, then there exists 1/x∈F s.t. x⋅1/x=1 • (D) The distributive law: x(y+z)=xy+xz, ∀x,y,z∈F$

Math 521 - 1/24

$Course Overview • The real number system • Metric spaces and basic topology • Sequences and series • Continuity • Topics from differential and integral calculus Grading Homework assignments 20% Quiz (Feb. 9) 5% Midterm 1 (Mar. 9) 20% Midterm 2 (Apr. 13) 20% Final (May. 10 7:45-9:45 AM) 35% A ≥90% B ≥80% C ≥70% D ≥60% Tutoring • Tom Stone @VV B205 • Monday 2:30 - 4:30 PM • Tuesday 2:00 - 4:00 PM What is Analysis • Proof • How calculus works • Fundamental axioms Number Systems • Natural Numbers: N={1,2,3,…} • Integers: Z={0,±1,±2,±3,…} • Rational Numbers: Q={a/b│a,b∈Zb≠0} • Real numbers R: fill the "holes" in the rational numbers √2 is not rational • There is no rational number p such that p^2=2 • Proof by contradiction • Assume there is a rational number p such that p^2=2 • Then p=m/n , where m,n∈Z, n≠0, and m,n have no common factor • (m/n)^2=2⇒m^2/n^2 =2⇒m^2=2n^2 • So m is even • m=2k (k∈Z⇒(2k)^2=2n^2⇒4k^2=2n^2⇒2k^2=n^2 • So n is also even • m,n are both division by 2 • This contradicts the fact that m,n have no common factor • So no such p exists$

Shawn Zhong

Shawn Zhong

Math 521

Math 521 - 1/31

Math 521 - 1/29

Math 521 - 1/26

Math 521 - 1/24

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