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Math4111
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Math4111 L3

Lecture 3

Review

Let S=ZS=\mathbb{Z}.

  1. Let E={xS:x>0,x2<5}E=\{x\in S:x>0,x^2<5\}. What are sup Esup\ E and inf Einf\ E?

    sup E=2,inf E=1sup\ E=2,inf\ E=1

  2. Can you find a subset ESE\subset S which is bounded above but not bounded below?

    E={xS:x<0}E=\{x\in S:x<0\}

  3. Does SS have the least upper bound property?

    Yes, ES\forall E\subset S that tis non-empty and bounded above, SupES\exist Sup E\in S.

  4. Does SS have the greatest lower bound property?

    Yes, ES\forall E\subset S that tis non-empty and bounded below, InfES\exist Inf E\in S.

Continue

LUBP

Proof that LUBP    GLBPLUBP\implies GLBP.

Let SS be an ordered set with LUBP. Let B<S be non-empty and bounded below.

Let L=yS:yL=y\in S:y is a lower bound of B}\}. From the picture, we expect sup L=inf Bsup\ L=inf\ B First we'll show sup Lsup\ L exists.

  1. To show LϕL\neq \phi.

    BB is bounded below     Lϕ\implies L\neq\phi

  2. To show LL id bounded above.

    BB is not empty     xB    x\implies \exists x\in B\implies x is a upper bound of LL.

  3. Since SS has the least upper bound property, supLsup L exists (in SS).

Let's say α=sup L\alpha=sup\ L. We claim that α=inf B\alpha=inf\ B. We need to show 22 things.

  1. To show α\alpha is a lower bound of BB, γB,αγ\forall \gamma\in B,\alpha\leq \gamma.

    Let γB\gamma\in B, then γ\gamma is an upper bound of LL.

    Since α\alpha is the least upper bound of LL, αγ\alpha\leq \gamma.

  2. To show α\alpha is the greatest lower bound of BB, β>α,β\forall \beta>\alpha,\beta is not a lower bound of BB.

    Let β>α\beta>\alpha. Since α\alpha is an upper bound of LL, βL\beta\notin L.

    By definition of LL, β\beta is not a lower bound of BB.

Thus α=inf B\alpha=inf\ B

Field

additionmultiplication
closure\checkmark\checkmark
commutativity\checkmark\checkmark
associativity\checkmark\checkmark
identity\checkmark (denoted 00)\checkmark (denoted 11)
inverses\checkmark (denoted x-x)\checkmark (exists when x0x\neq 0 denoted 1/x1/x or x1x^{-1})
distributivity\checkmark (distributive of multiplication over addition)

Examples: Q,R,C\mathbb{Q},\mathbb{R},\mathbb{C}

Non-examples: N\mathbb{N} fails A4,A5,M5, Z\mathbb{Z} fails M5

Another example of field: Z/5Z={1,2,3,4,5}\mathbb{Z}/5\mathbb{Z}=\{1,2,3,4,5\}, a,bZ/5Z\forall a,b\in \mathbb{Z}/5\mathbb{Z}, a+b=(a+b)mod5a+b=(a+b)\mod 5, ab=(ab)mod5a\cdot b=(a\cdot b)\mod 5

Some properties of fields: see Proposition 1.14,1.15,1.16

Remark:

  1. It's more helpful if you try to prove these yourselves. The proofs are "straightforward".
  2. For this course, it's not important to remember which properties are axioms, etc.

Example of proof:

1.14(a) x+y=x+z    y=zx+y=x+z\implies y=z

Proof:

x+y=x+zx+y=x+z,

(x)+(x+y)=(x)+(x+z)(-x)+(x+y)=(-x)+(x+z),

by A3, (x+x)+(y)=(x+x)+(z)(-x+x)+(y)=(-x+x)+(z),

0+y=0+z0+y=0+z,

y=zy=z.

Chain of equalities.

1.16(a) xF,0x=0\forall x\in \mathbb{F}, 0x=0

  1. A4, where 0 is defined.
  2. Since 00 is defined in the addition, identity. The proposition says something about multiplication by 0. The only proposition that relates the addition and multiplication is Distributive law.

0x=(0+0)x=0x+0x0x=(0+0)x=0x+0x, cancel 0x0x on both side we have 0x=00x=0.

Ordered Field (1.17)

An ordered field is a field FF which is also an ordered set, such that

  1. x+y<x+zx+y<x+z if x,y,Fx,y,\in F and y<zy<z,
  2. xy>0xy>0 if xF,yF,x>0x\in F,y\in F,x>0 and y>0y>0.

Prop 1.18

If x>0x>0 and y<zy<z, then xy<yzxy<yz.

Proof: y<z    0<zyy<z\implies 0<z-y, x(zy)>0    xz>xyx(z-y)>0\implies xz>xy

We define R\mathbb{R} to be the unique ordered field with LUBPLUBP. (The existence and uniqueness are discussed in the appendix of this chapter).

Theorem 1.20

  1. (Archimedean property) If x,yRx,y\in \mathbb{R} and x>0x>0, then nN\exists n\in \mathbb{N} such that nx>ynx>y.
  2. (Q\mathbb{Q} is dense in R\mathbb{R}) If x,yRx,y\in \mathbb{R} and x<yx<y, then \exists p\in \mathbb{Q}$$ such that x<p<y$.