Homework 6, Advanced Calculus 1

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Homework 6, Advanced Calculus 1

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1. Rudin Chapter 3 Exercise 24ab.

Solution: (a) For every Cauchy sequences {pn}, {qn}, we readily verify

• limn→∞d(p_n, p_n) = lim_n→∞0 = 0 ⇒ {p_n} ∼ {pn}.

• {p_n} ∼ {q_n} ⇒ 0 = lim_n→∞d(p_n, q_n) = lim_n→∞d(q_n, p_n) ⇒ {q_n} {p_n}.

• {p_n} ∼ {q_n} and {q_n} ∼ {r_n} ⇒ lim_n→∞d(p_n, r_n) ≤ lim_n→∞(d(p_n, q_n) + d(q_n, r_n)) = 0 + 0 = 0 ⇒ {p_n} ∼ {r_n}.

(b) Suppose {pn} ∼ {p⁰_n} and {qn} ∼ {q_n⁰}, we have limn→∞d(pn, p⁰_n) = limn→∞d(qn, q⁰_n) = 0.

Therefore

n→∞lim d(pn, qn) ≤ lim

n→∞d(pn, p⁰_n) + lim

n→∞d(p⁰_n, q_n⁰) + lim

n→∞d(q⁰_n, q_n⁰)

and the first, third terms are both zero due to their equivalence. So the function ∆ is independent of choice of representatives and therefore defines a function on X^∗. Since d is a metric, whose reflexivity and triangle inequality are preserved under limit operation, we see clearly that ∆ is a metric on X^∗. For the remaining parts of this problem, we will write P for both a Cauchy sequence and the equivalence class it belongs.

2. Rudin Chapter 3 Exercise 24c.

Solution: Let {P^m}m⊂ X^∗ be Cauchy (in ∆). For all  > 0, there are two  − N arguments we will utilize:

(i) Since {P^m}mis ∆− Cauchy, there exists M such that

∆(P^m, P^r) = lim

k→∞d(p^m_k, p^r_k) <  3. for all m, r > M_. The statement can be rephrased as

There exists M_, K_,m,n∈ N so that d(p^mk, p^r_k) < 

3 ∀m, r > M_, k > K_,m,n.

(ii) For all m ∈ N, since P^m= {p^m_n} is a Cauchy sequence in X, there exist N_mso that d(p^m_n, p^m_l ) <



3 for all n, l > Nm.

From the sequence of sequences above, take P = {p^k_N

k}k, where Nk is chosen according to (ii) above. We must show that P is Cauchy in d and ∆(P^m, P ) → 0 as m → ∞. Again starting with an arbitrary  > 0, for k > l > M in (i), we have

d(p^k_Nk, p^l_Nl) ≤ d(p^k_Nk, p^k_v) + d(p^k_v, p^l_v) + d(p^l_v, p^l_Nl)

for any v. Take v > max(Nk, Nl, Kk,l, then according to the descriptions of (i) and (ii), the sum above is less than . Therefore, P is Cauchy in d.

To show that P^m→ P , we estimate

∆(P^m, P ) = lim

k→∞d(p^m_k, p^k_Nk).

Follow similar logics, we have

d(p^m_k, p^k_N

k) ≤ d(p^m_k , p^m_v) + d(p^m_v , p^k_v) + d(p^k_v, p^k_N

k).

For m > M_, k > max(M_, N_m), and v > max(M_, N_m, N_k, K_,m,k), the above sum is then less than  and result follows.

3. Rudin Chapter 3 Exercise 24d.

Solution: With the given constant (and therefore Cauchy) sequences P_p and P_q, it is clear from definition that ∆(P_p, P_q) = d(p, q). The surjective mapping ϕ : X → X^∗ given by ϕ(p) = P_p is clearly injective. Since for all p 6= q in X, we have ∆(P_p.P_q) = d(p, q) > 0 and therefore P_p6= P_q.

4. Rudin Chapter 3 Exercise 24e.

Solution: Take any P ∈ X^∗ and  > 0, let the Cauchy sequence {pk} be a representative of P . Since it is Cauchy, there is N_ so that d(p_n, p_m) <  for all n, m > N_. We then have

∆(P_pN, P ) = lim

k→∞d(p_k, p_N) ≤ 

and therefore ϕ(X) is dense in X^∗. If X is complete, {p_k} is convergent with limit p. Therefore

∆(Pp, P ) = lim

k→∞d(p, pk) = 0 or Pp= P . Therefore ϕ(X) = X^∗.

5. Rudin Chapter 3 Exercise 21.

6. Rudin Chapter 3 Exercise 22.

7. Rudin Chapter 3 Exercise 23.

8. Given f : A → B and {Eα} a collection of subsets of B, prove that (a) f⁻¹(∪αEα) = ∪αf⁻¹(Eα).

(b) f⁻¹(∩αEα) = ∩αf⁻¹(Eα).

(c) f⁻¹(E_α^c) = f⁻¹(Eα)^c.

9. Prove that 8a is still true with f⁻¹ replaced by f , but 8b and 8c no longer hold.

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