Since 02− (n)2 = −n2≤ 1 for any n ≥ 1, an∈ K for all n ≥ 1

1. Quizz 6

(1) Prove that the subset K = {(x, y) ∈ R² : x²− y² ≤ 1} of R² is not sequentially compact by constructing a sequence in K which does not have any convergent sub- sequence in K.

Proof. Let an= (0, n) for n ≥ 1. Since 0²− (n)² = −n²≤ 1 for any n ≥ 1, a_n∈ K for all n ≥ 1. On the other hand, ka_nk = n for all n ≥ 1. For any subsequence (anj) of (an), we have kanjk = n_j ≥ j for all j ≥ 1. This shows that (a_nj) is also unbounded for all j ≥ 1 and thus (anj) is never convergent in R².

 (2) Let Y = {(x, y) ∈ R² : x²+ y² < 1} and (Y, d_Y) be the metric subspace of (R², d) defined by Y. Here d is the Euclidean metric on R². Prove that (Y, d_Y) is not complete by find a Cauchy sequence in (Y, dY) so that it is divergent in (Y, dY).

Proof. Let an= (1 − 1/n, 0) for any n ≥ 1. Then (1 − 1/n)²+ 0² < 1 for all n ≥ 1.

We find an∈ Y for all n ≥ 1. For any n > m ≥ N, dY(an, am) = kan− a_mk = 1

m − 1 n < 1

m.

For any  > 0, take N = [1/] + 1. For any n > m ≥ N, we have kan− a_mk < .

This implies that (a_n) is a Cauchy sequence in (Y, d_Y). Now let us show that (a_n) is divergent in (Y, dY).

Suppose not. Let a ∈ Y so that lim

n→∞a_n = a. Since a ∈ Y, d_Y(a, 0) = kak < 1.

Moreover, d(a_n, 0) = 1 − 1/n for any n ≥ 1. By triangle inequality, 0 ≤

kak −

 1 − 1

n

   

= |d_Y(a, 0) − d_Y(a_n, 0)| ≤ d_Y(a_n, a).

Since lim

n→∞an= a, we find lim

n→∞dY(an, a) = 0. By Sandwich principle,

n→∞lim    

kak −

 1 −1

n

   

= 0.

This implies that kak = 1 which leads to a contradiction to the assumption that kak < 1.

 (3) Let Y be a closed subset of R^p and (Y, d_Y) be the metric subspace of (R^p, d) defined

by Y. Prove that (Y, d_Y) is a complete metric space.

This is a special case of hw 7. (2)

(4) The n-th partial sum of an infinite seriesP∞

n=1a_n of real numbers is defined to be s_n=

n

X

i=1

a_i = a₁+ a₂+ · · · + a_n, for n ≥ 1.

1

2

The infinite series

∞

X

n=1

a_n is said to be convergent (in R) if the sequence of partial sums (s_n) is convergent in R. Suppose that there exists a sequence of positive real numbers (b_n) such that

(a) |an| ≤ b_nfor all n ≥ 1, and (b)

∞

X

n=1

bn is convergent.

Prove that

∞

X

n=1

a_n is absolutely convergent, i.e.

∞

X

n=1

|a_n| is convergent.

Proof. Let un= |a1| + · · · + |a_n| and t_n= b1+ · · · + bn. SinceP∞

n=1bnis convergent, (t_n) is a bounded sequence. Since |a_n| ≤ b_n for all n ≥ 1, 0 ≤ u_n≤ t_n for all n ≥ 1.

This implies that (un) is bounded. Since un+1 = un+ |an+1| ≥ u_n for any n ≥ 1, (un) is nondecreasing. By monotone sequence property, (un) is convergent.