0, there exists N ∈ N such that if n &gt

Advanced Calculus Study Guide 4 November 13, 2017

Sequences

Recall A sequence {p_n} in X is a function f : N → X such that f(n) = pn ∈ X.

Definition The sequence {pn} converges in X if there exists p ∈ X such that for each  > 0, there exists N ∈ N such that if n > N then d(p, pn) < . We say {p_n} converges to p. Or equivalently, p is the limit of {p_n}.

Notation p_n→ p ⇐⇒ lim

n→∞{p_n} = p.

Definitions

(a) If {pn} does not converge to any point in X then it diverges.

(b) The range of {p_n} is:

{x ∈ X|x = p_n for some n}.

(c) We say {p_n} isbounded in X if the range of {p_n} is bounded in X.

True/False Questions

(a) p_n → p, p_n → p⁰ ⇒ p = p⁰. True.

(b) {pn} is bounded ⇒ p_n converges. False.

(c) pn converges ⇒ {pn} is bounded. True.

(d) p_n → p ⇒ p is a limit point of the range of {p_n}. False.

(e) p is a limit point of E ⊆ X, then there exists some sequence {p_n} such that p_n→ p. True.

(f) p_n → p ⇐⇒ Every neighborhood of p contains all but finitely many terms in {p_n}. True.

Remark “All but finitely many” is equivalent to “Almost all.”

Proof

a) Let  > 0. Then there exists N₁ ∈ N such that n > N1, d(p, p_n) < 

2. Similarly there exists N₂ ∈ N such that n > N2, d(p⁰, p_n) < 

2. Then for n > max (N₁, N₂):

d(p, p⁰) ≤ d(p, p_n) + d(p⁰, p_n) < 2 2 = .

Thus d(p, p⁰) <  for all  > 0. Thus d(p, p⁰) = 0. Hence p = p⁰. b) Consider Ex. (4) where there is oscillation between two points.

c) Suppose p_n → p. Since 1 > 0 (our choice of ), there exists N ∈ N such that for n > N implies d(p_n, p) < 1. Let r = max (1, d(p, p₁), . . . , d(p, p_n)). Then p_i ∈ B(p, r) for all i ∈ N.

d) Consider Ex. (3) where the sequence is simply one point.

Advanced Calculus Sequences and Series Study Guide 4 (Continued) November 13, 2017 e) p ∈ E⁰. For all n ∈ N such that pn ∈ E such that d(p_n, p) < _n¹ and p_n 6= p. There for a sequence {p_n}. Let  > 0. Choose N > 1

. Then for each n > N , we have d(p, p_n) < 1 n < 1

N < .

Thus this sequences {p_n} converges to p as desired.

f) (⇒) Suppose p_n → p. Given B(p, ), there exists N ∈ N such that when n > N, it fol- lows that d(p, p_n) < , i.e. p_n∈ B(p, ), leaving only finitely many points, p₁ through p_npossible.

(⇐) For all  > 0, B(p, ) contains almost all {p_n}. For  > 0, let m = max{n ∈ N | pⁿ6∈ B(p, )}.

Then n > m implies p_n∈ B(p, ), i.e. d(p_n, p) < .

Limit Laws Let {s_n}, {T_n} be sequences in C and sn→ s and t_n→ t. Then the following hold:

• lim

n→∞(s_n+ t_n) = s + t.

• lim

n→∞(cs_n) = cs for all c ∈ C.

• lim

n→∞(c + s_n) = c + s for all c ∈ C.

• lim

n→∞(s_nt_n) = st for all c ∈ C.

• lim

n→∞

1 s_n = 1

s for all c ∈ C.

Proof a) Idea: 

|(s_n+ t_n) − (s + t)| ≤ |s_n−s|+|t_n−t|

. Let  > 0. Then there exists N₁ ∈ N such that n > N1 implies d(sn, s) < ₂^ and there exists N2 ∈ N such that n > N¹ implies d(tn, s) < ₂^. For n > max (N₁, N₂),

d(s_n+ t_n, s + t) = |(s_n+ t_n) − (s + t)|

= |sn− s| + |tn− t|

< .

b) Idea: |cs_n− cs| ≤ c|s_n− s|.

c) Idea: |s_nt_n− st| = |(s_n− s)(t_n− t) + s(t_n− t) + t(s_n− s)|. Let  > 0 Choose k = max(s, t, 1, ).

Then there exists N₁, N₂ such that n > N₁ implies d(s_n, s) < 

3k and similarly if n > N₂ implies d(t_n, t) < 

3k. Let N = max N₁, N₂ For n > N we know:

|s_nt_n− st| = |(s_n− s)(t_n− t) + s(t_n− t) + t(s_n− s)|

= ² 9k² + 

3 +  3

=  9k + 

3 +  3

< .

Theorem Let X be a metric space. Given a sequence {pn} in X, let E^∗ = {p ∈ X | ∃ a subsequence {p_ni} ⊂ {p_n} such that lim

i→∞p_ni = p}.

Page 2

Advanced Calculus Sequences and Series Study Guide 4 (Continued) November 13, 2017 Then E^∗ is closed, i.e. E^∗
0

⊆ E^∗.

Proof. Let E = {pn| n ∈ N} be the set of points determined by {pⁿ}.

Case 1: E = {q₁, . . . , q_m}, i.e. E contains finitely many points of X.

=⇒ E^∗ contains finite number of points

=⇒ E^∗
0

= ∅ ⊆ E^∗.

=⇒ E^∗ is closed.

Case 2: E contains infinitely many points of X.

=⇒ ∀ q ∈ E^∗
0

, ∃ p_n1 ∈ {p_n} such that p_n1 6= q. Let δ = d(q, p_n1) > 0.

Since q ∈ E^∗
0

=⇒ ∃ x₁ ∈ N_δ/2(q) ∩ E^∗\ {q}.

Also since x₁ ∈ E^∗ =⇒ ∃ n₂ > n₁ such that p_n2 ∈ N_δ/2(x₁) ∩ E.

=⇒ d(q, p_n2) ≤ d(q, x₁) + d(x₁, p_n2) < δ.

Similarly, ∃ x₂ ∈ N_δ/2²(q) ∩ E^∗\ {q}.

Also since x₂ ∈ E^∗ =⇒ ∃ n₃ > n₂ > n₁ such that p_n3 ∈ N_δ/2²(x₂) ∩ E.

=⇒ d(q, pn3) ≤ d(q, x2) + d(x2, pn3) < δ/2.

q p_n1

x₁ p_n2 δ

δ/2

δ/2

x₂ p_n3

δ/4 δ/4

For k ≥ 2, suppose pn1, . . . , pn_k are chosen such that d(pnj, q) < δ/2^j−2 for 2 ≤ j ≤ k, then we choose

x_k ∈ N_δ/2^k(q) ∩ E^∗\ {q} and p_nk+1 ∈ N_δ/2^k(x_k) ∩ E such that

d(q, p_nk+1) ≤ d(q, x_k) + d(x_k, p_nk+1) < δ/2^k−1.

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Advanced Calculus Sequences and Series Study Guide 4 (Continued) November 13, 2017

p_nk

q xk

pnk+1

δ/2^k−2

δ/2^k−1

δ/2^k

δ/2^k

Since lim

k→∞d(q, p_nk) ≤ lim

k→∞δ/2^k−2 = 0, {p_nk} is a subsequence of {p_n} converging to q. We have shown that if q ∈ E^∗
0

then q ∈ E^∗, i.e E^∗
0

⊆ E^∗ and E^∗ is closed.

Page 4