Midterm 1 Review

Using the axioms for multiplication from the definition of a field F, prove the following statements hold for all x,y,z ∈ F: (a) If x≠0 and xy=xz then y=z. (b) If x≠0 and xy=x then y=1. (c) If x≠0 and xy=1 then y=(1/x). (d) If x≠0 then 1/(1/x) = x.

(a) Axioms (M) from the definition of a field give: xy = xz => 1/x * (xy) = 1/x * (xz) => (1/x * x)y = (1/x * x)z => (x * 1/x)y = (x * 1/x)z => 1 * y = 1 * z => y = z. (b) Take z = 1 in (a). Then, we get xy = x * 1 => 1/x * (xy) = 1/x * (x) => (x * 1/x)y = (x * 1/x) => 1 * y = 1 => y = 1. (c) Take z = 1/x in (a). Then, we get xy = x * (1/x) => 1/x * (xy) = (x * 1/x) * 1/x => (x * 1/x)y = (x * 1/x)1/x => 1 * y = 1 * 1/x => y = 1/x. (d) Since 1/x * x = x * 1/x = 1, (c) (with 1/x in place of x and x in place of y) gives (d) as long as we have 1/x ≠ 0 which holds since using the addition axioms: 0x = (0 + 0)x = 0x + 0x => 0x + (-0x) = 0x + 0x + (-0x) => 0 = 0x, and from the multiplication axioms: 1/x * x = x * 1/x = 1 ≠ 0.

Given sets A, B, define A+B={a+b:a∈A and b∈B}. Follow these steps to prove that if A and B are nonempty and bounded above then sup(A + B) = supA + supB. (a) Let s=supA and t=supB. Show s+t is an upper bound for A+B. (b) Now let u be an arbitrary upper bound for A + B, and temporarily fix a∈A. Show t≤u−a. (c) Finally, show sup(A + B) = s + t.

(a) Let a + b ∈ A + B, where a ∈ A and b ∈ B. Then, a ≤ s and b ≤ t. => a + b ≤ s + t, so s + t is an upper bound for A + B. (b) We know that a + b ≤ u => b ≤ u - a, so u - a is an arbitrary upper bound for B. Recall from (a) that b ≤ t. Then, b ≤ u - a => b ≤ t ≤ u - a => t ≤ u - a, a true statement because t is a least upper bound. (c) From (b) we have that t ≤ u - a => a ≤ u - t. We also know from (a) that a ≤ s, so we get s ≤ u - t => s + t ≤ u. This means u is an arbitrary upper bound for A + B, and thus, by the definition of a least upper bound, s + t is one for A + B.

Let x_n ≥ 0 for all n ∈ 𝐍. (a) If (x_n) → 0, show that (√x_n)→0. (b) If (x_n) → x, show that (√x_n)→√x.

(a) Let ɛ > 0. Since x_n -> 0, there exists N ∈ 𝐍 s.t: | x_n - 0 | = | x_n | = x_n < ɛ^2, for n ≥ N. Then, for n ≥ N: | √x_n - 0 | = | √x_n | = √x_n < √ɛ^2 = ɛ. (b) If x = 0, then part (b) follows from part (a). So, suppose x > 0. Let ɛ > 0. Then: | √x_n - √x | = | ((√x_n - √x)(√x_n + √x))/(√x_n + √x) | = | (x_n - x)/(√x_n + √x) | ≤ | (x_n - x)/√x < ɛ <=> | x_n - x| < ɛ√x. Now x_n -> x, so there exists N ∈ 𝐍 s.t | x_n - x | < ɛ√x for n ≥ N. Hence, for n ɛ N: | √x_n - √x | < ɛ.

Decide whether the following propositions are true or false, providing a short justification for each conclusion. (b) If every proper subsequence of (x_n) converges, then (x_n) converges as well. (c) If (x_n) contains a divergent subsequence, then (x_n) diverges. (d) If (x_n) is bounded and diverges, then there exist two subsequences of (x_n) that converge to different limits.

(b) T: if (x_n) was convergent, every subsequence of (x_n is convergent. (c) T: by Bolzano-Weierstrass, there exists a convergent subsequence (x_n_k) and since (x_n) diverges we can obtain another subsequence which stays away from the limit of (x_n_k) and apply Bolzano-Weierstrass to this subsequence. (d) T: the convergent subsequence is monotone so must increase or decrease to its limit and thus (x_n) as a monotone sequence cannot get away from this limit.

Axioms of addition

1) if x in F and y in F, x + y = F 2) commutative: x+y = y+x for x, y in F 3) associative: (x+y) + z = x + (y+z) for x, y, x in F 4) F contains an element 0 s.t 0+x = x for all x in F 5) to every x in F corresponds a -x in F s.t x + (-z) = 0

axioms of multiplication

1) if x, y in F, xy in F 2) commutative: xy=yx for x,y in F 3) associative: (xy)z = x(yz) 4) F contains an element 1 ≠ 0 s.t 1x = x 5) if x ≠ 0, there exists an element 1/x s.t x*(1/x) = 1 6) distributive law: x(y+z) = xy+xz

Suppose that (x_n) is a sequence of real numbers. Then (pertaining to limsup and liminf):

1) limsup(x_n) = lim(y_n), y_n = sup{x_k: k ≥ n}. 2) liminf(x_n) = lim(z_n), z_n = inf{x_k: k ≥ n}

form of odd number

2k + 1 for some integer k

form of even number

2k for some integer k

Def of convergence of a sequence

A sequence (a_n) converges to a real number a if for every ɛ > 0, there exists an N ∈ 𝐍 s.t whenever n ≥ N, it follows that | a_n - a | < ɛ.

Order Limit Theorem for Sequences

Assume (a_n), (b_n) converge to limits a, b. Then, (a) if a_n ≥ 0, then a ≥ 0. (b) if a_n ≤ b_n, then a ≤ b. (c) if there exists c in R s.t c ≤ b_n, then c ≤ b.

Prove √3 is irrational.

By way of contradiction, suppose there exists a rational number whose square is 3. Then, ∃ p, q ∈ Z s.t: (1) (p/q)^2 = 3. We may also assume that p, q have no common factor, since if they did, we could cancel it out and rewrite the fraction in its lowest terms. So, (1) implies: (2) p^2 = 3q^2. Then, p^2 is divisible by 3, which means that p must be divisible by 3. => p = 3r, where r is an integer. Then, using (2) we have: 9r^2 = 3q^2 => 3r^2 = q^2. This implies that q^2 is divisible by 3, which means that q must be divisible by 3. Thus, we have shown that p, q are both divisible by 3 when they were assumed to have no common factor, a contradiction.

Show that there is no rational number r satisfying 2^r = 3.

Case i): If r > 0, then r = p/q with p, q ∈ Z and we may also assume p, q > 0 since we could cancel a negative sign if they were both negative. Then, 2^p/q = 3 => 2^p = 3^q, a contradiction since 2^p is even and 3^q is odd. Case ii): If r=0, then 2^0 = 3 = 1 = 3, a contradiction. Case iii): If r < 0, then r = p/q with p, q ∈ Z and we may also assume p < 0, q > 0 since we could move the negative sign to p if p > 0, q < 0. Then: 2^p/q = 3 => 2^p = 3^q, a contradiction since 2^p < 1 for p < 0 and 3^q > 1 for q >0. Thus, in all possible cases we have a contradiction and so there is no rational number r satisfying 2^r = 3.

convergent sequence <=>

Cauchy

An ordered set has the least upper bound property if

E, a subset of S, is non-empty and bounded above, then sup(E) exists in S.

Theorem 2.3.2

Every convergent sequence is bounded.

T/F? Provide example if false to show statement doesn't hold: A∩(B∪C)=(A∩B)∪C.

False: consider the sets A = {1}, B = {2}, C= {3}. Then, A∩(B∪C) = Ø and (A∩B)∪C = {3}, which are evidently not equal.

If s1 = √2, and s_n+1 = √(2 + √sn), (n = 1, 2, 3, ...), prove that s_n converges. (Hint: Use the Monotone Convergence Theorem. To show s_n is bounded from above, use induction to show that s_ n ≤ 2 for n = 1, 2, 3, ... ).

First note {s_n} is increasing since: {s_n} = {√2, √(2 + √2), ...} and the square root is an increasing function. We claim s_n ≤ 2 and prove this using induction: n = 1: s_n = √2 ≤ 2. Inductive hypothesis: We suppose s_n ≤ 2 for some n ∈ 𝐍. Then: s_n+1 = √(2 + √s_n) ≤ √(2 + √2) ≤ √4 = 2. So s_n+1 ≤ 2, and hence by induction, s_n ≤ 2 for all n ∈ 𝐍. Therefore, {s_n} is increasing and bounded from above, and thus converges by the Monotone Convergence Theorem.

Thm 2.10.4

If (x_n)

Theorem 2.2.8

If a sequence converges, its limit is unique.

Monotone Convergence Theorem

If a sequence is monotone and bounded, then it converges.

Binomial Thm

If x, y in R and n in 𝐍, then: (x+y)^n = sumfrom 0 to n (n choose k)*(x^n-k)*y^k

Let (a_n) and (b_n) be Cauchy sequences. Prove that c_n = |a_n −b_n| is a Cauchy sequence directly from the definition.

Let ɛ > 0. Since (a_n) and (b_n) are Cauchy sequences, there exists N_a, N_b ∈ 𝐍 s.t: | a_n - a_m | < ɛ/2 for n, m ≥ N_a and | b_n - b_m | < ɛ/2 for n, m ≥ N_b. Take N = max(N_a, N_b). Then, or n, m ≥ N: | c_n - c_m | = | |a_m - b_n | - | a_m - b_m || ≤ | (a_n - b_n) - (a_m - b_m) | (by reverse triangle inequality) = | (a_n - a_m) + (b_m - b_n) | ≤ | a_n - a_m | + | b_m - b_n | (by triangle inequality) = | a_n - a_m | + | b_n - b_m | < ɛ/2 + ɛ/2 = ɛ.

Show that if x_n ≤ y_n ≤ z_n for all n ∈ 𝐍, and if lim(x_n) = lim(z_n) = l, then lim y_n = l as well.

Let ɛ > 0. Since lim(x_n) = lim(z_n) = l, there exists N_x, N_z ∈ 𝐍 s.t: | x_n - l | < ɛ for n ≥ N_x, | z_n - l | < ɛ for n ≥ N_z. Ig N = max(N_x, N_z), then for all n ≥ N: -ɛ < x_n - l ≤ y_n - l ≤ z_n - l < ɛ => -ɛ y_n - l < ɛ => |y_n - l | < ɛ. So, lim y_n = l.

Prove ||a| − |b|| ≤ |a − b|. (The unremarkable identity a = a − b + b may be useful.)

Note: |a | = | a - b + b | ≤ | a - b | + | b | => | a | - | b | ≤ | a - b | and | b | = | b - a + a | ≤ | b - a | + | a | => -| a - b | ≤ | a | - | b |. Then: -| a - b | ≤ | a | - | b | ≤ | a - b | => | | a | - | b | | ≤ | a - b |.

Converse of P -> Q

Q -> P

Let (a_n) be a bounded (not necessarily convergent) sequence, and assume lim b_n = 0. Show that lim(a _n * b_n) = 0. Why are we not allowed to use the Algebraic Limit Theorem to prove this?

Since a_n is bounded, there exists M > 0 s.t | a_n | ≤ M. Let ɛ > 0. Since lim(b_n) = 0, there exists N ∈ 𝐍 s.t n ≥ N => | b_n | < ɛ/M. Then, for n ≥ N: | a_n * b_n | = | a_n | * | b_n | ≤ M | b_n | < M (ɛ/M) = ɛ, so lim(a_n*b_n) = 0. We are not allowed to use the Algebraic Limit Theorem to prove this because (a_n) is not necessarily convergent.

Algebraic Limit Theorem for Sequences

Suppose (s_n), (t_n) are sequences and converge to limits s, t. Then, (a) lim(s_n + t_n) = s+t (b) lim(c * s_n) = c*s (c) lim(s_n * t_n) = s*t (d) lim(1 / s_n) = 1/s (s.t s is non-zero)

Let (x_n) and (y_n) be given, and define (z_n) to be the "shuffled" sequence (x1,y1,x2,y2,x3,y3,...,xn,yn,...). Prove that (z_n) is convergent if and only if (x_n) and (y_n) are both convergent with lim x_n = lim y_n.

Suppose (z_n) is convergent. Then, note (x_n) and (y_n) are subsequences of (z_n) and so by Theorem 2.4.2, (x_n) and (y_n) are convergent sequences with the same limit as (z_n) so that lim(x_n) = lim(y_n). Now suppose (x_n) and (y_n) are both convergent with lim(x_n) = lim(y_n) = l. Let ɛ > 0. Then, there exists N_x, N_y ∈ 𝐍 s.t: n ≥ N_x => | x_n - l | < ɛ n ≥ N_y => | y_n - l | < ɛ. Let N = max(N_x, N_y). Then, for n ≥ N: | z_n - l | = { | z_2m-1 - l | if n = 2m-1 is odd { | z_2m - l | if n = 2m is even = { | x_m - l | if n = 2m-1 is odd { | y_m - l | if n = 2m is even < { ɛ if n = 2m-1 is odd since n ≥ N => 2m-1 ≥ 2N_x-1 => m ≥ N_x { ɛ if n = 2m is even since n ≥ N => 2m ≥ 2N_y => m ≥ N_y. In either case, | z_n - l | < ɛ.

If r is rational (non-zero) and x is irrational, prove r+x is irrational.

Suppose by way of contradiction that r+x is rational. Then, r+x = p/q, where p, q ∈ Z, q ≠ 0. => x = p/q - r. But r ∈ Q, so -r ∈ Q and x = p/q - r ∈ Q/ But we assumed x to be irrational, so we have a contradiction. Thus, r+x is irrational.

Prove that convergence of {sn} implies convergence of {|sn|}.

Suppose s_n -> s. Let ɛ > 0. Then, there exists n ∈ 𝐍 s.t for n ≥ N, | s_n - s | < ɛ. Then, for n ≥ N: | | s_n | - | s | | ≤ | s_n - s | < ɛ (by reverse triangle inequality) => | s_n | -> | s |, that is, {| s_n |} converges.

T/F? liminf(x_n) ≤ limsup(x_n)

T

Let E be a nonempty subset of an ordered set; suppose α is a lower bound of E and β is an upper bound of E. Prove that α ≤ β.

We know E is bounded below, which implies that α ≤ e for all e ∈ E. We also know that it is bounded above, i.e, that e ≤ β for all e ∈ E. => α ≤ e and e ≤ β => α ≤ e ≤ β => α ≤ β.

Let A be a nonempty set of real numbers which is bounded below. Define the set −A to be the set of all numbers −x, where x ∈ A. Prove that inf A = − sup(−A).

We show -sup(-A) is the greatest lower bound for A and hence inf(A) = -sup(-a): Lower bound: Let x ∈ A. Then, -x ∈ -A. Thus, -x ≤ sup(-a) => -sup(-A) ≤ x. Greatest lower bound: Let z be a lower bound for A. Then -z is an upper bound for -A since if -x ∈ -A where x ∈ A then z ≤ x => -x ≤ -z. Hence, sup(-A) ≤ -z => z ≤ -sup(=A).

Let A ⊆ R be nonempty and bounded above, and let c∈R. This time define the set cA={ca:a∈A}. If c ≥ 0, show that sup(cA) = csupA.

We show csup(A) is the least upper bound for cA and hence sup(cA) = csup(A). Upper bound: Let x ∈ cA. Then, x = ca for some a ∈ A. Thus, a ≤ sup(A) => ca ≤ csup(A) => x ≤ csup(A). Least upper bound: Let z be an upper bound for cA. Then for all a ∈ A, ca ≤ z. Now suppose c > 0. Then, we have a ≤ z/c for all a ∈ A. Then, z/c is an upper bound for A. Thus, sup(A) ≤ a/c => csup(A) ≤ z.

Cauchy sequences => convergent sequences are...

bounded!

Use a similar strategy to the one in Example 2.5.3 to show lim b^(1∕n) exists for all b ≥ 0 and find the value of the limit. (The results in Exercise 2.3.1 may be assumed).

do on paper

Where does sequence as n -> infinity (1 + 1/n)^n converge?

e

Let a_n = (n^n)/n! . Show that lim n→∞ (an+1)/an= e. (Hint: You may use that limn→∞ (1 + 1n)^n = e).

easy

Bolzano-Weierstass Thm

every bounded sequence contains a conv subsequence

If (x_n) is a sequence, then it conv to infinity if

for every M in R, there exists a N in 𝐍 s.t x_n > M for all n ≥ N. (....there exists a N in 𝐍 s.t x_n < M for all n ≥ N. ....there exists a N in for -infinity).

Corollary 2.4.4

if a sequence has subsequences that conv to different limits, then the sequence diverges

a sequence is monotone if

it is either increasing (a_n ≤ a_n+1) or decreasing (a_n ≥ a_n+1)

A sequence (x_n) of real # conv iff....

liminf(x_n) = limsup(x_n) = x are finite and equal, in which case lim(x_n) = x.

Let A and B be subsets of R. (a) If x ∈ (A∩B)c, explain why x ∈ Ac ∪Bc. This shows that (A∩B)c ⊆ Ac ∪Bc. (b) Prove the reverse inclusion (A ∩ B)c ⊇ Ac ∪ Bc, and conclude that (A ∩ B)c = Ac ∪ Bc. (c) Show (A ∪ B)c = Ac ∩ Bc by demonstrating inclusion both ways.

meh

Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit: (a) lim (2n+1)/(5n+4) = 2/5 = 0. (b) lim (2n^2)/(n^3+3) = 0. (c) lim (sin(n^2 ))/(n^1/3) = 0.

on paper

Extend the result proved in Example 2.5.3 to the case | b | < 1; that is, show lim(b^n) = 0 if and only if − 1 < b < 1.

paper

Given two sequences {an} and {bn} bounded below, prove that: lim sup n→∞ (an + bn) ≤lim sup n→∞ an + lim sup n→∞ bn.

paper

Between any two real numbers is a ___ number.

rational

Thm 2.4.2

subsequences of a convergent sequence converge to the same lim as the original sequence

A sequence (x_n) is bounded if...

there exists a number M > 0 s.t | x_n | ≤ M for all n ∈ 𝐍.

infimum (greatest lower bound)

this is the greatest lower bound s.t every element a of the set is bounded below by it (s.t inf ≤ a), but any other arbitrary lower bound x is s.t (x ≤ inf)

supremum (least upper bound)

this is the smallest upper bound s.t every element a of the set is bounded above by it (s.t a ≤ sup), but any other arbitrary upper bound x is s.t (sup ≤ x)

A sequence (a_n) is called a Cauchy sequence if for every ɛ > 0, there exists N in 𝐍 s.t:

whenever m, n ≥ N, | a_n - a_m | < ɛ.

liminf/sup(x_n) = +/- infinity if

y_n -> +/- infinity, z_n -> +/- infinity

Triangle inequality

| a + b | ≤ | a | + | b |

Contrapositive of P -> Q

~Q -> ~P