MATH410 Exam 1
The completeness axiom
If S is a nonempty set bounded above, then among the set of upper bounds for S. There is a least upper bound
For S = [0, inf), is S sequentially compact?
No! we can define {an} = {n}. This does not converge as it is not bounded.
Archimedean Property
i) for positive number c, there is a natural number n such that n > c ii) for positive number ε, there is a natural number n such that 1/n < ε
A set S of real numbers is inductive if:
i) the number 1 is in S ii) if the number x is in S, the number x + 1 is also in S
Binomial Formula
(a+b)^n = (sum from 0 to n) (n choose k)*a^(n-k)*b^k
Extreme Value Theorem
A continuous function on a closed bounded interval, f:[a, b] -> R attains both a min and a max value
Monotone Convergence Theorem (what does a monotonically increasing sequence converge to? decreasing?)
A monotone sequence converges if and only if it is bounded. A monotonically increasing sequence converges to the supremum. A monotonically decreasing sequence converges to the infimum.
Definition of a sequence
A real-valued function whose domain is a set of natural numbers
Definition of convergence
A sequence {an} converges to a number a provided that for every positive number ε there is an index N such that |an - a| < ε for all indices n ≥ N
How to define boundedness
A set S is bounded above and below provided there is a nonnegative number M such that: |x| ≤ M for all x in S
Alternate term of denseness (in terms of limits and sequences)
A set S is dense in R if and only if every number x is the limit of a sequence in S
What does dense in R mean
A set S is dense in R provided every interval I = (a, b) where a < b contains a member of S
Sequential compactness definition
A set of real numbers S is sequentially compact provided that every sequence {an} in S has a subsequence that converges to a point that belongs to S.
Definition of a closed set
A subset S of R is said to be closed provided that if {an} is a sequence in S that converges to a number a, then a also belongs in S.
Why does every sequence have a monotone subsequence?
Because of peak indices. There are either only finitely many peak indices, or infinitely many.
Why does every bounded sequence have a convergent subsequence?
By the theorem that every sequence has a monotone subsequence, we know that that this subsequence is bounded, and thus converges.
How to prove monotone convergence theorem
Define c as the supremum. We claim {an} converges to c. let epsilon > 0. |an - c| < epsilon for all indicies n >= N Thus: c - epsilon < an < c + epsilon for all indices n >=N since the number c is an upper bound for the set S, we have: an <= c < c + epsilon for ever index n. Since l is a least upper bound, since {an} is monotonically increasing and by the completeness axiom: c - epsilon , aN <= an for all indices n >=N Thus, {an} converges to c
How to prove that [a, b] is closed
Define some {cn} in [a,b] that converges to c. lim[b - cn] = b -c Since cn belongs in [a, b] b - cn >= 0 define {dn} = {b - cn} that converges to d such that {dn} is non-negative Suppose d < 0 let epsilon = -d/2 we can see that (d - epsilon, d + epsilon) consists of entirely negative values, thus no part of {dn} belongs in the interval, so c must be less than or equal to b A similar argument can be made for c - a >= 0
How to prove the Quotient Property of convergent sequences
Do it using the product property of convergent sequences combined with the reciprocal lemma
Sequential Density of Rationals
Every number is the limit of a sequence of rational numbers (as the set of rational numbers is dense in R)
Proof Monotone Subsequence Theorem
Every sequence has a monotone subsequence. Case 1: there are finitely many peak indices. We can choose an index N such that there are no peak indices greater than N. Now define monotonically increasing sequence of {an} define n1 = N+1. Now suppose k is an index such that positive integers: n1 < n2 < ... < nk have been chosen such that an1 < an2 < ... < ank Since nk > N, the index nk is not a peak index. Hence there is an index nk+1 > nk such that ank+1 > ank. We thus defined a strictly increasing sequence of positive integers {nk} having the property that the subsequence {ank} is strictly increasing Case 2: there are infinitely many peak indices. for each natural number k, let nk be the kth peak index. Directly from the definition of peak index, it follows that the subsequence {ank} is monotonically decreasing.
Triangle Inequality
For any pair of a and b: |a + b| ≤ |a| + |b| -|a| - |b| ≤ a+b ≤ |a| + |b|
Linearity Property of Convergent Sequences
For any two alpha, beta: lim(alpha*an + beta*bn) = alpha*lim(an) + beta*lim(bn)
Nested Interval Theorem
For each natural number n, let an, and bn be numbers such that an < bn and consider In = [an, bn] assume: In+1 is in In for every index n. Also assume lim[an - bn] = 0. Then there is exacltly one point x that belongs to In for all n, and both {an} and {bn} converge to this point. This is because {an} monotonically increases, and {bn} monotonically decreases.
Principle of Mathematical Induction
For each natural number, mathematical assertation S(n) is true if whenever k is in N, S(k) is true and S(k+1) is true
Product Property of Convergent Sequences
If {an} converges to a, and {bn} converges to b, then: lim[anbn] = lim[an]*lim[bn] = a*b
Comparison Lemma of convergent sequences
If {an} converges to a, and {bn} converges to b, there is a nonnegative number C and an index N1 such that |bn - b| ≤ C*|an - a| for all indices n ≥ N1
Polynomial Property of Convergent Sequences
If {an} converges to a, for a polynomial p: R-> R: lim(p(an)) = p(a)
Reciprical Property of Convergent Sequences
If {bn} converges to b, then {1/bn} converges to 1/b
Image Interval Theorem
Let I be an interval and suppose that the function f: I->R is continuous. Then its image f(I) also is an interval. Proved by the IVT
Sequential Compactness Theorem
Let a, b be numbers such that a < b. Then every sequence in [a, b] is sequentially compact as it has a subsequence that converges to a point in [a, b]
How to prove the Sum Property of Convergent Sequences
Let epsilon > 0, we must find an index N such that: |(an+bn) - (a+b)| < epsilon Taking it equal to |(an-a)+(bn-b)|, we can apply the triangle inequality such that: |(an-a)+(bn-b)| <= |an - a| + |bn - b| Since {an} converges to a and epsilon/2 is positive, we can choose an index N1 such that |an - a| < epsilon/2 all n>=N1 same for b, but use N2 Take N = max{N1, N2}, thus: |(an+bn) - (a+b)| <= |an-a| + |bn-b| < epsilon/2 + epsilon/2 = epsilon
Property of the limit of subsequences
Let {an} converge to a. Then, every subsequence of {an} also converges to a.
Is the set of rational numbers closed?
No! Since there is a sequence of rational numbers that converges to an irrational number (sqrt(2)) by the denseness of rational numbers.
Is S = (0, 2] sequentially compact?
No! as {1/n} is a sequence in S that does not converge to a point ini S.
How to prove the sequential compactness theorem
Prove that [a, b] has a convergent subsequence, prove that this limit belongs in [a, b], then every sequence {xn} is bounded, and we can find a monotone subsequence of such {xnk} that converges to a point in [a, b] Show that [a, b] is closed. Use the monotone subsequence theorem to show that there is a monotone subsequence. Then use the monotone convergence theorem to show that it converges.
How to prove that if I is an interval that if f: I->R is continuous, then image f(I) is also an interval (a continuous function defined on [a, b] is bounded)
Show that f(I) is a convex set. Let y1, y2 be points in f(I) such that y1 < y2. We must show that [y1, y2] is contained in f(I). Let y1 < c < y2. Since y1, y2 are in f(I), there are points x1 and x2 in I with f(x1) = y1 and f(x2) = y2. Let J be the closed interval having x1 and x2 as endpoints., then J is contained in I since the set I is an interval and is thus convex. Thus we can use the IVT for the function f: J->R in order to conclude there is an x0 in J in which f(x0) = c. So x0 belongs in I and f(x0) = c. Thus [y1, y2] is contained in f(I)
How do we prove the Intermediate Value Theorem
The Bisection Method (a product of nested interval theorem) Assume f(a) < c and f(b) > c. Let a1 = a and b1 = b. For a natural number n, suppose the interval [an, bn] contained in [a, b] has been defined such that f(an) ≤ c and f(bn) > c Consider midpoint mn = (an+bn)/2 If f(mn) ≤ c, define an+1 = mn and bn+1 = bn If f(mn) > c define an+1 = an and bn+1 = mn bn+1 - an+1 = (bn - an)/2 So that (bn - an) = (b - a)/2^(n-1) Thus by the Nested Interval Theorem, there is an x0 in [a, b] to which both {an} and {bn} converge as {f(an)} and {f(bn)} converge to f(x0) = c
Intermediate Value Theorem
The function f:[a, b] -> R is continuous. Let c be a number strictly between f(a) and f(b) such that f(a) < c < f(b) or f(b) < c < f(a) then there is a point x0 in (a, b) such that f(x0) = c
Distribution of integers
There is a unique integer in the interval [c, c+1) for c as any number. There is also no integer in the interval (k, k+1) where k is an integer
True or false: every convergent sequence is bounded
True as |an| = |(an - a) + a| ≤ |an-a| + |a| |an| ≤ 1 + |a| letting epsilon= 1. Thus define M = max of 1 + |a|, ... |an| such that |an| ≤ M
How to prove extreme value theorem
Use the Completeness Axiom to define a supremum. define natural number n such that c - 1/n must be less than some value of f(x) by the completeness axiom such that c - 1/n < f(xn) ≤ c for every index n The sequential compactness theorem says that a subsequence {xnk} of {xn} converges to a point x0 in [a, b]. {f(xnk)} is a subsequence of {f(xn)} that converges to c so c = f(x0) and x0 is the maximizer of f. then do the same for -f to prove that there is a minimum.
How to prove the product property of convergent sequences
We must find lim(anbn - ab) = 0 take alphan = an - a and betan = bn - b lim alphan and betan = 0 anbn - ab = (a+alphan)*(b+betan) - ab = a*betan + b*alphan + alphan*betan taking the limit equals zero By the lemma that the product of two sequences with one converging to zero converges to zero
Monotonically decreasing definition
a(n+1) ≤ an for every index n
Difference of powers
a^n - b^n = (a-b)*(a^(n-1)+a^(n-2)b + ... + b^(n-1)) = (a-b)*(sum from 0 to n-1 of a^(n-1-k)b^k
Monotonically increasing definition
an ≤ a(n+1) for every index n
Definition of maximum value (maximizer) of image f(D)
f(x) ≤ f(x0) for all x in D. A nonempty set has a maximum provided that the set is bounded above and contains its supremum.
Definition of Continuity
f: D->R is said to be continuous at point x0 in D provided whenever {xn} is a sequence in D that converges to x0, the image {f(xn)} converges to f(x0). lim(n->inf) f(xn) = f(x0)
Quotient Property of Convergent Sequences
if {an} converges to a and {bn} converges to b, then: lim(an/bn) = a/b
Sum, product, and quotient property of continuous functions
lim[(f+g)(xn)] = f(x0) + g(x0) lim[(fg)(xn)] = f(x0)*g(x0) lim[(f/g)(xn)] = f(x0)/g(x0)
Sum property of convergent sequences
lim[an + bn] = lim(an) + lim(bn) Proved by triangle inequality.
Closed sets theorem of convergent sequences [a, b]
{cn} is a sequence in [a, b]. If {cn} converges to a number c, then c also belongs in [a, b]