Chapter 2 Convergence And Sequences Flashcards
Prop 2.3 a in X (xn →a, d) if and only if
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.
Prop 2.8
proof
unique limit
*****
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.
Prop 2.12
convergence of subsequence
.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 → ∞.
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
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
DEF 2.1 limit
B(a, ε).
d(xn
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.
EXAMPLE
Show that 1/√n → 0 in R.
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.
in R we have limit properties:
Algebra of limits
(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.
in R we have limit properties:
sandwich rule
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.
in R we have limit properties:
following sequences tend to 0
(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).
Example 2.6. Show that
(1/√n,(n−1)/n)
→ (0, 1) in R^2 with the Euclidean metric.
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).
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.
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).
Proposition 2.8
unique limit
A sequence (xn) in a metric space (X, d) has at most one limit.
Proposition 2.9
under d1 sequence in R^m converges
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_∞
DEF 2.11 SUBSEQUENCE
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.
PROP 2.12 for subsequences
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.
fn to converge to a function f∈ C[a, b]:
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
POINTWISE CONVERGENCE
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.
Example 2.14. Let (f_n) be the sequence of functions defined by
f_n(x) = 1 + x +x^2/2! + · · · + x^n/n!
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
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∞.
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]).
Proposition 2.15 pointwise.
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.
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?
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
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].
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].
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∞).
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.
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)
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.
