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Lemma 1 : (∀ x in F) x * 0 = 0
- Use Distributive Law
- Additive inverse of (x*0) is -(x * 0)
Lemma 2 : (∀ x in F) -x = -1 * x
First step: x + ( -1 * x ) = (x * 1) + ( x * -1 )
with (x * 1) being the multiplicative identity and the second part by commutativity.
- Use distributive law with x * (1 + -1)
- Prove RHS = 0, then add (-x) to both sides.
Lemma 3 : (-1)2 = 1
Distributive Law with a = -1, b = -1, c = 1
Lemma 4 : (∀ x in F) If x does not = 0, then x does not equal -x
Consider trichotomy, so case 1 (x<0) and case 2 (x>0)
+ x
Use transitivity x + x < x < 0
Lemma 5 : (∀ x in F) x2 >= 0
Trichotomy. 3 cases: x = 0, x > 0, x < 0
For case 1 and 2 : *x for both sides.
For case 3 : add -x, so that you can * -x as -x > 0
Lemma 6 : Archimedian Property v1 (the set of natural numbers are unbounded)
Proof by Contradiction. Assume N is bounded above.
So it has an upper bound x, so x-1 is not an upper bound. So can pick n>x-1 to exist. so n+1>x false.
Lemma 7 : Archimedean Property v2. 1/n → 0
Definition of convergence.
N>1/ε so 1/N<ε
Lemma 8 : Every sequence contains a monotone subsequence.
2 cases: infinitely many peaks & finitely many peaks
Case 1: infinitely many peaks so ank >= ank+1 so monotone decreasing
Case 2 : Let ak be the last peak, so an1 < an2 so strictly monotone increasing
Proposition 1 : Every rational number has a finite decimal expansion or an infinite recurring decimal expansion.
Let p/q be in the rational….
10n1p / q = a1 + r1/q
remainder = 0 then expansion terminates and is finite
remainder repeats: all subsequent digits repeat
Proposition 2
Root 2 is not rational.
Proposition 3 : A sequence (an) is bounded if and only if (∃C in R)(∀n in N) |an| <= C
=> Definition of bounded with L and U, set C = max{ |L|, |U| }.
<= Set L = -C and U = C
Proposition 4 : If an is a convergent sequence, then (an) is bounded.
Set ε = 1 and so a - ε < an < a + ε
Consider terms nefore N1 so (∀n<= N1 -1) m <= an <= M
(∃m,M in R)(∀n in N)(min { a-1 , m} <= an <= max {a+1, M} , and set the min = L and max = U
Proposition 5 : an → a <=> every subsequence ank → a
=> Convergence def. All indices are strictly increasing, so nk >= N so replace ank with an.
<= Suppose every subsequence converges to a, set k = nk, so replaces ank = ak
Theorem 1 : Triangle inequality
Write |x| and |y| using expanded bounds and combine
Theorem 2 : Algebraic Limit Theorem
Algebraic Limit Theorem (an → a & bn → b)
can → ca
an + bn → a+b
anbn → ab
an/bn → a/b
Theorem 2 : Algebraic Limit Theorem part a (can → ca)
ε / |c|
(∀ ε / |c| >0 )… where |an - a| < ε / |c|
Multiply the c to the other side
Theorem 2 : Algebraic Limit Theorem part b (an + bn → a+b)
ε / 2
Take Na and Nb and M = max (Na, Nb)
Use triangle inequality
Theorem 2 : Algebraic Limit Theorem part c (anbn → ab)
Take Na and Nb and M = max (Na, Nb)
|anbn - ab| = |anbn - abn + abn -ab| < …
Use bn is bounded so |bn| <= C and set C into where that is
Make each term < ε / 2 so +1 on denominator
Theorem 2 : Algebraic Limit Theorem part d (an/bn → a/b)
Prove (1/bn → 1/b) then apply part c
0 < C < |b|
|bn - b| = |b - bn| < C
Try to express bn in terms of something else
Theorem 3 : Sequence Squeeze Theorem ((∀n in N)(an <= xn <= bn) and an → L and bn → L then, xn → L
Express the expanded L - ε < … < L + ε
Take Na and Nb and M = max (Na, Nb)
Combine using transitivity
Theorem 4: Order Limit Theorem
((∀n in N)(an >= 0) => a >= 0
((∀n in N)(an <= bn) => a<=b
((∀c in R)(((∀n in N) c<=bn) => c<= b)
Theorem 4: Order Limit Theorem part a : ((∀n in N)(an >= 0) => a >= 0
Proof by contradiction : suppose a < 0
|an - a| < |a|. => an < 0
Theorem 4: Order Limit Theorem part b : ((∀n in N)(an <= bn) => a<=b
Consider the sequence (bn - an) → (b - a)
Use part a so b - a >= 0
Theorem 4: Order Limit Theorem part c : ((∀c in R)(((∀n in N) c<=bn) => c<= b)
Define bn = c (the constant sequence) and apply part b
Theorem 5 : Monotone Convergence Theorem
Monotone + Bounded => Convergence
Theorem 5 : Monotone Convergence Theorem : if monotone + bounded => Convergence
2 Cases: increasing case and decreasing case
E.g. Monotone increasing case
By def of bounded, there exists sup = s
s - ε < aN <= an <= s < s + ε by transitivity it converges to the sup
Theorem 6 : Bolzano - Weierstrass Theorem
Every bounded sequence contains a convergent subsequence.
(by Lemme 8 we know every sequence contains a monotone subsequence, so if its bounded, then by monotone convergence theorem, it converges)
Def 1 : Partial Ordering
Reflexivity : a<=a
Antisymmetry : a<=b and b<=a, then a=b
Transitivity : a<=b and b<=c , then a<=c
Def 2 : Total Ordering
If it is a partial ordering and any two elements are comparable ( a<=b or b<=a).
Def 3 : Maximal element
No element x is greater than the maximal element m.
Def 4 : Greatest element
(∀s in S) s<= g and g is in S
Def 5 : Field
A set F with two operations + and *, that satisfy the 11 axioms.
Def 6 : Ordered Field
A field with total ordering <=
01) If x <= y , then x+z <= y+z
02) If x <= y , then x*z <= y*z
Def 7 : Upper Bound
(∀s in S) s<= B
Def 8 : Supremum (least upper bound)
1) (∀s in S) s<= B
B is an upper bound
2) (∀y < B)(∃a in S) a>y
No element less than B is an upper bound
Def 9 : Lower Bound
(∀s in S) s >= b
Def 10 : Infimum (greatest lower bound)
1) (∀s in S) s>= b
b is a lower bound
2) (∀y > b)(∃a in S) a<y
No element greater than b is lower bound
Def 11 : Least Upper Bound Property
If a subset is non empty and is bounded above, then there exists a supremum in the set.
Def 12 : Greatest Lower Bound Property
If a subset is non empty and is bounded below, then there exists an infimum in the set.
Def 13 : Real Numbers
An ordered field with the least upper bound property.
Def 14 : Absolute value
|x| = x , if x>=0
|x|= -x , if x<0
Def 15 : Sequence
A function a : N → R
Def 16 : Convergence of an → L
(∀ε >0)(∃N in N)(∀n>=N) |an - L| < ε
Def 17 : ε neighbourhood
For a in R & ε >0. Vε (a) = { x in R : |x - a| < ε}
(points whose distance from a is less than ε)
Def 18 : Convergence (topological)
an → a if every ε neighbourhood of a contains all terms of the sequence from some point on. Contains finite terms.
(∀ε >0)(∃N in N)(∀n>=N) an is in Vε (a)
Def 19 : Bounded
(an) is bounded if (∃L,U in R)(∀n in N) L <= an <= U
Def 20 : Monotone…
Monotone increasing if (∀n in N) an <= an+1
Strictly monotone decreasing is (∀n in N) an > an+1
Def 21 : Partial Sum and Series
The mth partial sum of (bn) is = b1 + b2 + b3 …
Series is the sequence of partial sums. ie. b1 , b1+b2 , b1+b2+b3, …
If the partial sum converges to B, then the series equals B.
Def 22 : Subsequence
A subsequence of (an) is (ank). Where the index are strictly monotone increasing.
Def 23 : Peak
am is a peak of a sequence (an) if (∀n > m) am >= an