4.5 KiB
Lecture 3
Review
Let S=\mathbb{Z}.
-
Let
E=\{x\in S:x>0,x^2<5\}. What aresup\ Eandinf\ E?sup\ E=2,inf\ E=1 -
Can you find a subset
E\subset Swhich is bounded above but not bounded below?E=\{x\in S:x<0\} -
Does
Shave the least upper bound property?Yes,
\forall E\subset Sthat tis non-empty and bounded above,\exist Sup E\in S. -
Does
Shave the greatest lower bound property?Yes,
\forall E\subset Sthat tis non-empty and bounded below,\exist Inf E\in S.
Continue
LUBP
Proof that LUBP\implies GLBP.
Let S be an ordered set with LUBP. Let B<S be non-empty and bounded below.
Let L=y\in S:y is a lower bound of B$}$. From the picture, we expect sup\ L=inf\ B First we'll show sup\ L exists.
-
To show
L\neq \phi.Bis bounded below\implies L\neq\phi -
To show
Lid bounded above.Bis not empty\implies \exists x\in B\implies xis a upper bound ofL. -
Since
Shas the least upper bound property,sup Lexists (inS).
Let's say \alpha=sup\ L. We claim that \alpha=inf\ B. We need to show 2 things.
-
To show
\alphais a lower bound ofB,\forall \gamma\in B,\alpha\leq \gamma.Let
\gamma\in B, then\gammais an upper bound ofL.Since
\alphais the least upper bound ofL,\alpha\leq \gamma. -
To show
\alphais the greatest lower bound ofB,\forall \beta>\alpha,\betais not a lower bound ofB.Let
\beta>\alpha. Since\alphais an upper bound ofL,\beta\notin L.By definition of
L,\betais not a lower bound ofB.
Thus \alpha=inf\ B
Field
| addition | multiplication | |
|---|---|---|
| closure | \checkmark |
\checkmark |
| commutativity | \checkmark |
\checkmark |
| associativity | \checkmark |
\checkmark |
| identity | \checkmark (denoted 0) |
\checkmark (denoted 1) |
| inverses | \checkmark (denoted -x) |
\checkmark (exists when x\neq 0 denoted 1/x or x^{-1}) |
| distributivity | \checkmark (distributive of multiplication over addition) |
Examples: \mathbb{Q},\mathbb{R},\mathbb{C}
Non-examples: \mathbb{N} fails A4,A5,M5, \mathbb{Z} fails M5
Another example of field: \mathbb{Z}/5\mathbb{Z}=\{1,2,3,4,5\}, \forall a,b\in \mathbb{Z}/5\mathbb{Z}, a+b=(a+b)\mod 5, a\cdot b=(a\cdot b)\mod 5
Some properties of fields: see Proposition 1.14,1.15,1.16
Remark:
- It's more helpful if you try to prove these yourselves. The proofs are "straightforward".
- For this course, it's not important to remember which properties are axioms, etc.
Example of proof:
1.14(a) x+y=x+z\implies y=z
Proof:
x+y=x+z,
(-x)+(x+y)=(-x)+(x+z),
by A3, (-x+x)+(y)=(-x+x)+(z),
0+y=0+z,
y=z.
Chain of equalities.
1.16(a) \forall x\in \mathbb{F}, 0x=0
- A4, where 0 is defined.
- Since
0is 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+0x, cancel 0x on both side we have 0x=0.
Ordered Field (1.17)
An ordered field is a field F which is also an ordered set, such that
x+y<x+zifx,y,\in Fandy<z,xy>0ifx\in F,y\in F,x>0andy>0.
Prop 1.18
If x>0 and y<z, then xy<yz.
Proof: y<z\implies 0<z-y, x(z-y)>0\implies xz>xy
We define \mathbb{R} to be the unique ordered field with LUBP. (The existence and uniqueness are discussed in the appendix of this chapter).
Theorem 1.20
- (Archimedean property) If
x,y\in \mathbb{R}andx>0, then\exists n\in \mathbb{N}such thatnx>y. - (
\mathbb{Q}is dense in\mathbb{R}) Ifx,y\in \mathbb{R}andx<y, then\exists p\in \mathbb{Q}such thatx<p<y.