0 so that kank Rk+1 ≤ M for any n ≥ 1

1. hw 7

(1) Assuming that the Bolzano-Weierstrass Theorem is true for R¹, prove that the Bolzano-Weierstrass Theorem is true for R^p for any p ∈ N by mathematical In- duction. (Hint: Let (a_n) be a bounded sequence in R^p+1. Write a_n= (x_n, y_n) for all n ≥ 1 where (xn) is a sequence in R^p and (yn) is a sequence in R. Show that (xn) is bounded in R^p and (yn) is bounded in R.)

Proof. We have proved the case when p = 1. Let us assume that the Bolzano- Weierstrass Theorem holds for p = k. Let (an) be a bounded sequence in R^k+1. Then there exists M > 0 so that ka_nk

R^k+1 ≤ M for any n ≥ 1. For each n ≥ 1, let us write a_n= (x_n, y_n) where x_n∈ R^k and y_n∈ R for n ≥ 1. For any n ≥ 1,

|y_n|, kx_nk

R^k ≤ ka_nk

R^k+1 = q

kx_nk²

R^k+ |y_n|² ≤ M.

This shows that (x_n) is a bounded sequence in R^kand (y_n) is a bounded sequence in R. By Bolzano-Weierstrass Theorem for R^k, (xn) has a subsequence (xnj) which is convergent to some point x in R^k. Since (yn) is bounded in R, (ynj) is also bounded in R. By Bolzano-Weierstrass Theorem for R, (ynj) has a further subsequence (y_njl) which converges to some y in R. Since (xnj) converges to x in R^k, its subsequence (x_njl) also converges to x. Now, the subsequence (a_njl) of (a_n) is convergent to (x, y). We prove that the Bolzano-Weierstrass Theorem is true for p = k + 1. By Mathematical induction, Bolzano-Weierstrass Theorem holds for all p ∈ N.  (2) Let (M, d) be a complete metric space and N be a subset of M. Let (N, dN) be the metric subspace of (M, d) associated with N. Prove that (N, d_N) is complete if and only if N is a closed subset of M.

Proof. Let us assume that N is a closed subset of M. We want to show that (N, d_N) is complete. Let (a_n) be a Cauchy sequence in (N, d_N). Then for any  > 0, there exists N∈ N so that dN(an, am) <  whenever n, m ≥ N. Since (N, dN) is a metric subspace of (M, d), d_N(x, x⁰) = d(x, x⁰) for any x, x⁰ ∈ N. For any n, m ≥ N_,

d(an, am) = dN(an, am) < .

This shows that (an) is a Cauchy sequence in (M, d). Since (M, d) is complete, (an) is convergent to some point a in (M, d). For the  > 0 given, there exists N_ ∈ N so that d(an, a) <  for any n ≥ N_. Since a_n ∈ N for any n ≥ 1, this shows that a is an adherent point of N. Since N is closed, a ∈ N. Since an, a ∈ N, d_N(a_n, a) = d(a_n, a) <  whenever n ≥ N_. This shows that (a_n) is convergent to a in (N, dN). Therefore (N, dN) is complete.

Let us assume that (N, d_N) is complete. We want to show that N is a closed subset of (M, d). In other words, let us prove that N contains all of its adherent points. Let a be an adherent point of N. Then there exists a sequence (an) in N so that (a_n) is convergent to a in (M, d), i.e. for any  > 0, there exists N_⁰ ∈ N so that d(an, a) < /2 whenever n ≥ N. Therefore d(an, am) <  whenever n, m ≥ N_⁰ in other words, (a_n) is a Cauchy sequence in (M, d). Since a_n ∈ N for any n ≥ 1, dN(an, am) = d(an, am) <  whenever n, m ≥ N_⁰. This implies that (an) is a Cauchy sequence in (N, d_N). By completeness of (N, d_N), (a_n) is convergent to some b ∈ N.

For the  given above, we can find N_⁰⁰ such that dN(an, b) < /2 whenever n ≥ N_⁰⁰.

1

2

Let N = max{N_⁰, N_⁰⁰}. For any n ≥ N_,

d(a, b) ≤ d(a_n, a) + d(a_n, b) = d(a_n, a) + d_N(a_n, b) <  2 + 

2 = .

This shows that d(a, b) <  for any  > 0. Hence d(a, b) = 0. Since d is a metric on M, a = b. We find that a = b ∈ N. Hence N is closed.

 (3) In Calculus, we have learned the following important properties for convergent se-

quences:

Theorem 1.1. Let (an) and (bn) be sequence of real numbers and k be a real number. Suppose that

n→∞lim an= a, lim

n→∞bn= b.

Then (a) lim

n→∞(a_n+ b_n) = a + b;

(b) lim

n→∞(ka_n) = ka;

(c) lim

n→∞anbn= ab (d) lim

n→∞

an

bn

= a

b if b ≥ 0.

In class, we also proved

Theorem 1.2. Let (a_n) be a sequence in R^p with a_n = (a¹_n, · · · , a^pn) for n ≥ 1.

Then (a_n) is convergent in R^p if and only if (a¹_n), · · · , (a^p_n) are convergent in R. In this case,

n→∞lim a_n=

n→∞lim a¹_n, · · · , lim

n→∞a^p_n .

Let (x_n) and (y_n) be sequences in R^p such that lim_n→∞x_n= x andlim_n→∞y_n= y. Use Theorem 1.1 and Theorem 1.2 to prove the following statements:

(a) lim

n→∞(x_n+ y_n) = x + y.

(b) lim

n→∞kx_n= kx.

(c) lim

n→∞anxn= ax.

(d) lim

n→∞

xn

b_n = x

b if b ≥ 0.

(e) lim

n→∞hx_n, y_ni = hx, yi.

Proof. Routine Check.