Adv Calc Test 1

Theorem: If 2|x²,

2|x. reverse also true.

De Morgan’s Laws

for sets or logical statements. (A^B)^c=A^cvB^c. (AvB)^c=A^c^B^c.

compliment

superscript c. basically not for sets. so the complement of a set is all the things not in the set.

Triangle Inequality

|a+b|<=|a|+|b|.

Two real numbers a and b are equal iff

for every real number ϵ > 0 it follows that |a - b| < ϵ

Axiom of Completeness

Every nonempty set of real numbers that is bounded above has a least upper bound (sup). Similarly, every nonempty set of real numbers that is bounded below has a most lower bound (inf).

sup A

The least upper bound of A.

inf A

the least lower bound of A

Prove something is an inf/sup

Bounded Above (definition)

Set A (which is a subset of R) is bounded above if there exists a real number, b, such that a <= b for all aϵA.

Bounded Below (definition)

Set A (which is a subset of R) is bounded below if there exists a real number, b, such that a >= b for all aϵA.

Least upper bound definition

A real number, s, is the least upper bound for a set A (which is a subset of R) if it 1) s is an upper bound for A and 2) if b is any upper bound for A, then s <= b.

Maximum definition

A real number a₀ is a maximum of the set A if a₀ is an element of A and a₀ >= a for all aϵA.

Minimum definition

A real number a₀ is a minimum of the set A if a₀ is an element of A and a₀ <= a for all aϵA.

Lemma: Assume sϵR is an upper bound for a set A (which is a subset of R). s = sup of A iff

for every choice of ϵ > 0, there exists an element a ϵ A satisfying s - ϵ < a.

Nested Interval Property

For each n ϵ N, assume we are given a closed interval I_n = [a_n,b_n] = {x ϵ R : a_n <= x <= b_n }. Assume also that each I_n contains I_n+1. Then, the resulting nested sequence of closed intervals has a nonempty intersection; that is the infinite intersection of I_n is non empty.

Archimedean Property

Given any number x ϵ R, there exists an n ϵ N satisfying n > x. Given any real number y > 0, there exists an n ϵ N satisfying 1/n < y.

Density of Q in R

For every two real numbers a and b with a < b, there exists a rational number r satisfying a < r < b.

Prove something is dense in something else.

density corollary. Given any two real numbers a < b,

there exists an irrational number t satisfying a < t < b.

A function f:A→B is one-to-one if

a₁≠a₂ in A implies that f(a₁)≠f(a₂) in B.

The function f is onto if,

given any bϵB, it is possible to find an element aϵA for which f(a)=b.

A ~ B

If there exists f: A → B that is 1-1 and onto, then set A has the same cardinality as B. This is written as A ~ B.

Set A is countable if

N ~ A.

If A is a subset of B and B is countable,

then A is either countable or finite.

If A₁, A₂, …, A_m are each countable sets,

then the union of these guys is countable.

If A_n is a countable set for each n ϵ N,

then the infinite union of the sets is countable.

diagonalization

Cantor’s Theorem

Given any set A, there does not exist a function f : A → P(A) that is onto

A sequence is

a function whose domain is N

A sequence converges to a real number a if

for every positive number ϵ, there exists an k ϵ N such that whenever n >= k it follows that |a_n - a| < ϵ.

Process for convergence of a sequence

Say ϵ > 0. Do scratch work to find k. State what k is. Show when n >= k, it follows that |a_n - a| < ϵ

Uniqueness of Limits

The limit of a sequence, when it exists, must be unique.

Algebraic Limit Theorem. Let lim a_n = a and lim b_n = b. Then,

lim(ca_n) = ca, lim(a_n + b_n) = a + b, lim(a_nb_n) = ab, lim(a_n/b_n) = a/b, provided b≠0.

Order Limit Theorem. Assume lim a_n = a and lim b_n = b.

If a_n >= 0 for all n ϵ N, then a >= 0. If a_n <= b_n for all n ϵ N, then a <= b. If there exists c ϵ R for which c <= b_n for all n ϵ N, then c <= b. Similarly if c >=

A sequence (a_n) is increasing if

a_n <= a_n+1 for all n ϵ N

A sequence (a_n) is decreasing if

a_n >= a_n+1 for all n A sequence (a_n) is increasing if N.

A sequence is monotone if

it is either increasing or decreasing

If a sequence is monotone and bounded,

then it converges