Chapter 2 Convergence And Sequences

Question 1

Prop 2.3
  a in X (xn →a, d) if and only if 

Answer 1

Let (x_n) be a sequence in a metric space X and

let a ∈ X. Then x_n → a in (X, d) if and only if d(x_n, a) → 0 in R.

Question 2

Prop 2.8
proof
unique limit
*****

Answer 2

Suppose that x_n → a and x_n → b.
By axiom M3,
0 ≤ d(a, b) ≤ d(a, x_n) + d(x_n, b)
for all n.

Now since x_n → a and x_n → b, d(a, x_n) → 0 and d(x_n, b) →0.

By the algebra of limits (2.5), d(a, xn) + d(xn, b) → 0 and, by the Sandwich Rule, the constant sequence d(a, b) → 0.

Hence d(a, b) = 0, and, by Axiom M1, a = b.

Question 3

Prop 2.12

convergence of subsequence

Answer 3

.Let x_n_1, x_n_2, x_n_3, . . . be any subsequence of (x_n). For any
ε > 0, there is a natural number N such that if n > N, d(x_n, a) < ε.
Now choose the natural number K so that n_K > N. If k > K, then
n_k > n_K > N, and we conclude that d(x_n_k, a) < ε. Thus (x_n_k) → a as n → ∞.

Question 4

Prop 2.3
Let (xn) be a sequence in a metric space X and
let a ∈ X. Then xn → a in (X, d) if and only if d(xn, a) → 0 in R.

*****PROOF

Answer 4

As d(x_n, a) ≥ 0,
d(x_n, a) ∈ B_R(0, ε ) ⇔ d(x_n, a) < ε ⇔ x_n ∈ B_X(a, ε).

By the definition of convergence, x_n → a in X if and only if d(x_n, a) →
0 in R.—

Given ε bigger than 0 there exists N in N st if n bigger than N |x_n -a| less than ε
⇔
d(x_n,a) = |x_n -a| → 0, |d(x_n,a)| less than ε
⇔
limit as n tends to infinity of x_n =a
limit as n tends to infinity of |d(x_n,a)|=0

Question 5

DEF 2.1 limit

B(a, ε).

d(xn

Answer 5

Let (x_n) be a sequence in the metric space X =
(X, d). Let a ∈ X. We say that x_n → a, or that (x_n) has limit
a ∈ X, or that (x_n) converges to a, if, given ε > 0, there exists
N ∈ N so that, for all n > N, x_n ∈ B(a, ε), that is, d(xn, a) < ε.

Question 6

EXAMPLE

Show that 1/√n → 0 in R.

Answer 6

Let ε > 0. By the Archimedean property of R, there exists
N ∈ N so that N > 1/ε^2

For n > N,
d(1/√n, 0) = 1/√n < 1/ √N < ε,
so 1/√n → 0.

Question 7

in R we have limit properties:

Algebra of limits

Answer 7

(a) if y_n → y and c ∈ R then cy_n → cy. In particular, if
yn → 0 and c ∈ R then y_n → 0

(b) if y_n → y and z_n → z then y_n + z_n → y + z. In particular,
if y_n → 0 and z_n → 0 then y_n + z_n → 0.

(c) if y_n → y and z_n → z then y_nz_n → yz. In particular, if
y_n → 0 and z_n → z then y_nz_n → 0.

(d) if y_n → y, z_n → z, each z_n ≠0 and z ≠ 0 then y_n/z_n →y/z.

Question 8

in R we have limit properties:

sandwich rule

Answer 8

If w_n ≤ y_n ≤ z_n for all n and w_n →l and
z_n → l then y_n → l.

In particular, taking w_n = 0 = l, if 0 ≤ y_n ≤ z_n and z_n → 0 then y_n → 0.

Question 9

in R we have limit properties:

following sequences tend to 0

Answer 9

(a) 1/n^p (= n^−p) for any p > 0. This includes 1/n, 1/n^2 and
1/√n.

(b) a^n for any a such that −1 < a < 1
(only 0 ≤ a < 1 will be
relevant here).

Question 10

Example 2.6. Show that
(1/√n,(n−1)/n)
→ (0, 1) in R^2 with the Euclidean metric.

Answer 10

d( (1/√n,(n−1)/n) , (0,1))

  \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_
√((1/√n− 0)^2 + (((n-1)/n) -1)^2)
      \_\_\_\_\_\_\_\_\_\_
=  √((1/n)  + (1/n^2))
≤
(√(2/n))  = √2 (1/√n)

By Summary 2.5(iii)(a) and (i)(a), √2 (1/√n) → 0. By the Sandwich
Rule,
d( (1/√n,(n−1)/n) ,  (0,1))
→ 0
so, by Proposition 2.3, 
(1/√n,(n−1)/n)
→ (0, 1).

Question 11

Example 2.7. Show that the sequence (f_n) in C[0, 1] defined by
f_n : [0, 1] → R, x → x^n/n,
converges to the constant function f(x) = 0 in C[0, 1] under both
d∞ and d1.

Answer 11

d_∞(f_n, f) = sup {|f_n(x) − f(x)| : x ∈ [0, 1]} = sup {x^n/n : x ∈ [0, 1]} = 1/n

as f_n increasing on [0,1], 1/n → 0 so, by Proposition
2.3, f_n → 0.

d_1(fn, f) = ∫ over [0,1] of |x^n/n -0|.dx
= ∫ over [0,1] of x^n/n .dx
= 1/(n(n+1)) less than 1/n^2

We know that 1/n^2 → 0 so, by the Sandwich Rule, d_1(f_n, f)→ 0 in R
and hence f_n → f in (C[0, 1], d_1).

Question 12

Proposition 2.8

unique limit

Answer 12

A sequence (xn) in a metric space (X, d) has at
most one limit.

Question 13

Proposition 2.9

under d1 sequence in R^m converges

Answer 13

Let (x_n) = (x^1_n, . . . , x^m_n) be a sequence in R^m and
let x = (x^1, . . . , x^m) ∈ R^m.

Under the metric d_1, x_n → x, as n → ∞ if and only if x^j_n → x^j ∈ R , as n → ∞ for all 1 ≤ j ≤ m

–
holds for d_2 and d∞ because a sequence in R^m converges under d_1 if and only if it converges under
d_2 if and only if it converges under d_∞

Question 14

DEF 2.11 SUBSEQUENCE

Answer 14

Let n_1 < n_2 < n_3 < . . . be an increasing
sequence of natural numbers and let (x_n) = x_1, x_2, x_3, . .

be a sequence in a metric space (X, d). Taking x_n_1, x_n_2, x_n_3, . . ., we
get a new sequence (x_n_k), which is a subsequence of (x_n).

For example we might have x_1, x_3, x_5, . . . or x_1, x_3, x_6, x_7, x_12, . . .. Note that
n_k > k.

Question 15

PROP 2.12 for subsequences

Answer 15

Let (x_n) be a sequence in a metric space (X, d)
converging to a limit a ∈ X. Then any subsequence (x_n_k) also converges
to a.

Question 16

fn to converge to a function f∈ C[a, b]:

Answer 16

fn to converge to a function f∈ C[a, b]:

• fn → f in the metric space (C[a, b], d1).
or
• fn → f in the metric space (C[a, b], d∞).

sequences (f_n(x)) for each x

Question 17

POINTWISE CONVERGENCE

Answer 17

We say that (f_n) converges pointwise to f if, for

each x ∈ [a, b], the sequence (f_n(x)) converges to f(x) in R.

Question 18

Example 2.14. Let (f_n) be the sequence of functions defined by
f_n(x) = 1 + x +x^2/2! + · · · + x^n/n!

Answer 18

for any x, e^x =Σ from i =0 to ∞ of (x^i/i!)

ie for each x,nΣ from i =0 to ∞ of (x^i/i!)→ e^x as n → ∞, and this exactly means that
(f_n) converges pointwise to e^x

Question 19

Let (f_n) be the sequence of functions defined by
f_n(x) = 1 + x +x^2/2! + · · · + x^n/n!

Show that, for b > 0, (fn) converges to e^x
in C([0, b]) under d∞.

Answer 19

Here |f_n(x) − e^x| = e^x − f_n(x). This has derivative e^x −f_{n−1}(x) = x^n/n! + x^{n+1}/(n+1)! +· · · ≥ 0 for all x ∈ [0, b] so it is increasing and
must take its maximum value at b. So d_∞(f_n(x), e^x) = |f_n(b)−e^b| →0 as f_n → f pointwise. Therefore (f_n) converges to e^x in C([0, b]).

Question 20

Proposition 2.15 pointwise.

Answer 20

If f_n converges to f in (C[a, b], d_∞), then f_n

converges to f in (C[a, b], d_1) and also f_n converges to f pointwise.

Question 21

Example 2.16.
In the space C([0, 1]), let f_n(x) =x^n

Does
the sequence (f_n) converge in any of the senses we’ve described?

Answer 21

Sketch the first few functions in the sequence. looks as if the graphs are getting closer and
closer together, and approaching some limit graph. limit graph seems to be
f(x)=
{0 if 0 ≤ x < 1
{1 if x = 1.
and this isn’t continuous. It turns out that whether this sequence
converges or not depends on our definition of convergence

Question 22

Example 2.16.
In the space C([0, 1]), let f_n(x) =x^n
show that:
(i)(fn) does not converge pointwise to any continuous function
f ∈ C[0, 1].

Answer 22

For x ∈ [0, 1], fn(x) = x^n
, which converges to 0 if x < 1, and converges to 1 if x = 1. Therefore if f_n converges pointwise to a function f ∈ C[0, 1], then

f(x)=
{0 if 0 ≤ x < 1
{1 if x = 1.
But this isn’t continuous, f ∉ C[0, 1].

Question 23

Example 2.16.
In the space C([0, 1]), let f_n(x) =x^n
show that:

(ii) (fn) does not converge in (C[0, 1], d∞).

Answer 23

from Proposition 2.15 that if (fn) converged in (C[0, 1], d∞),
then it would converge pointwise. But we know it doesn’t converge
pointwise.

Question 24

Example 2.16.
In the space C([0, 1]), let f_n(x) =x^n
show that:

(iii) (fn) converges to the zero function g(x) = 0 in (C[0, 1], d1)

Answer 24

For all n,
d1(fn, g) = 
∫ over[0,1] of  |fn(x) − g(x)|dx 
=
∫ over[0,1] of [x^n] dx 
=
1/{n + 1}  <  1/n

We know that 1/n → 0 so, by the Sandwich Rule, d_1(fn, g) → 0
and, by Proposition 2.3, f_n → g.

Question 25

function spaces and relationship between metrics

Answer 25

two natural
metrics on function spaces like C[0, 1] have very different properties.
When we deal with function spaces, we will try to be very careful
always to specify the metric.