1.30 Convergence and continuity

Question 1

Simple sandwich rule for limits

Answer 1

w_n ≤ x_n ≤y_n for all n
If w_n tends to 0 and y_n tends to 0 then x_n does

e.g. 0 ≤ d(x,y) can be used

Question 2

Archimedean property of reals

Answer 2

for any two positive real numbers a and b, a < b there exists a positive integer n such that na > b

Question 3

DEF 1.32 Convergence

Answer 3

We say x_n → x ( xn converges to x) if d(x_n, x) → 0 as n → ∞.

x_n → x if for any ε > 0 ∃N ∈ N st ∀n > N we
have d(x, x_ n) < ε.

Question 4

THM 1.33 convergence equivalence

Answer 4

Let (x_n) be a sequence in a metric space (X, d). Then the following
are equivalent:

1. x_n → x;
2. for every open U with x ∈ U, ∃N > 0 s.t. (n > N) ⇒ x_n ∈ U;
3. for every ε > 0 ∃N > 0 s.t. (n > N) ⇒ x_n ∈ B_ε(x).

Question 5

PROOF of THM 1.33 convergence equivalence

Answer 5

1→2: If x_n → x and x ∈ U, then there is a ball Bε(x) ⊂ U, since U is open. x_n → x so d(xn, x) < ε for n sufficiently large, so., x_n ∈ U for n sufficiently large
(there is N= N( ε) st ∀n>N lie in nhd Bε(x))
2→3: As U is open for each x ∃ε>0 s.tBε(x) ⊂ U. Thus x_n ∈Bε(x) as d(x,x_n)<ε
3→1: For a given ε>0 and large n x_n∈Bε(x) implies d(x_n,x)< ε

Question 6

THM 1.34 x ∈ s̅ when sequence…

Answer 6

Let S be a subset of the metric space X.
Then x ∈ s̅ IFF there is a sequence (xn) of points of S with x_n → x

Question 7

Corollary 1.35 (Closedness under taking limits)

Answer 7

A subset Y ⊂ X of a metric space
(X, d) is closed if and only if for every sequence (xn) in Y that is convergent in X its
limit is also in Y .

the closure s̅ is obtained from S by adding all possible limit points of sequences in S

Question 8

Def (f_n) converges pointwise

Answer 8

(f_n) converges pointwise to f if for each x in [a.b] (f_n(x)) converges to f(x) in R

f_n→ f pointwise as n tends to infitiy f_n(x) → f(x) for all x in [a,b]

Question 9

Proof: thm1.35 sequence and s̅

Answer 9

→
If x ∈ s̅ then by thm 1.28 we have V∩S ≠Ø for all nhds V of x. For each n we choose B_{1/n}(x). Then B_{1/n} ∩S ≠Ø.

For a sequence (x_n)∈S choose such a sequence in x_n ∈ B_{1/n}∩S.
as n→∞ d(x_n,x)→0 so x_n →x

←
If x ∉s̅ then there is a NHD U of x with U ∩S =Ø → no sequence in S can get into U so it cannot converge to x.

Question 10

Example:
Take (R^2, d1), where d_1(x, y) = |x_1−y_1|+|x_2−y_2|

Consider sequence
(1/n,{2n+1}/{n+1} ). Limit?

Answer 10

We guess its limit is (0, 2). To
see if this is right, look at
d_1((1/n,{2n+1}/{n+1} ),(0, 2))
= |1/n| + |* -2|
= (1/n) + (1/(n+1)) → 0 as n → ∞. So the limit is (0, 2).

Question 11

In C[0, 1] let f_n(t) = t^n and f(t) = 0 for 0 ≤ t ≤ 1.
Does fn → f in d1

Answer 11

d1(fn, f) = ∫_[0,1] t^n dt =
1/{n + 1} → 0
as n → ∞. So f_n → f in d1

graph is 0 then suddenly 1, looks more and more lik right side of a square

Question 12

In C[0, 1] let f_n(t) = t^n and f(t) = 0 for 0 ≤ t ≤ 1.
Does fn → f, in d∞?

Answer 12

d∞(fn, f) = max{t^n: 0 ≤ t ≤ 1} = 1 does not converge to 0 as n → ∞. So fn does not converge f in d∞

Question 13

In C[0, 1] let f_n(t) = t^n and
f(x) =
{ 0 for 0 ≤ x < 1.
{1 for x=1
Does fn → f, pointwise on [0,1]

Answer 13

f_n converges to f pointwise on [0,1] but f is not in X[0,1] as it is not continuous at 1

Question 14

Consider the discrete metric, when does a sequence converge in this metric?

Answer 14

d_0(x, y) =
{1 if x ≠ y,
{0 if x = y

Then x_n→ x ⇐⇒ d_0(x_n,x) → 0
d_0 is 1 or 0 so IFF d_0(x_n,x) =0 for sufficiently large n. If there is an n_0 st x_n=x for all n ≥ n_0

ie
All convergent sequences in this metric are eventually constant. So, for example
d0(1/n, 0) does not converge to 0.

Question 15

EXAMPLE pointwise convergence
f_n = 1 + x + x^2/2 +…+x^n/(n!) to

Answer 15

to f(x) = e^x

Question 16

Proposition 1.37. R^2 convergence with standard metrics a-n b_n

Answer 16

Take R^2 with any of the metrics d1, d2 and d∞. Then a sequence
xn = (an, bn) converges to x = (a, b) if and only if an → a and bn → b.

A similar result holds for R^m in general.

Question 17

PROOF
Proposition 1.37. R^2 convergence with standard metrics

Answer 17

Proof
If a_n → a and b_n → b then for any
ε > 0 there are N_a and N_b such that for N > N_a we
have |a_n − a| < ε/2 and for n > N_b |bn − b| < ε/2.

Thus for any n > N = max(Na, Nb):
ε > |an − a| + |bn − b| = d1(xn, x) ≥ d2(xn, x) ≥ d∞(xn, x),
which shows the convergence in all three metrics

converse: assume contradiction a_n does not converge to a: ie assume ∃ε >0 st for any N ∃ n>N s.t. |an − a| >ε. Then
d_1(x_n,x) ≥ d2(xn, x),≥ d∞(xn, x), = max {|a_n − a|,|b_n − b|} > |a_n − a| >ε showing divergence in three metrics

Question 18

For any x_n and x metrics inequality

Answer 18

d_1(x_n,x) ≥ d_2(x_n,x) ≥ d_∞(x_n,x)

Question 19

Thm 1.39 strength of convergence d_1

Answer 19

If f_n → f in (C[a, b], d∞), then f_n → f in (C[a, b], d1).
Informally speaking, d∞ convergence is stronger than d1 convergence.

Question 20

PROOF
Thm 1.38. strength of convergence d_1

Answer 20

PROOF

Question 21

REMARK of thm 1.38 point wise

Answer 21

d_∞ m.s and also f_n→ f pointwise

Question 22

DEF 1.40 continuity at a point

Answer 22

Let f : (X, dX) → (Y, dY ) be a map between metric
spaces. We say that f is continuous at x ∈ X if for each ε > 0 there is a δ_{ε,x} > 0 such
that d_Y (f(x′), f(x)) < ε for all x′ ∈ X whenever d_X(x′, x) < δ_{ε,x}.

Question 23

Another def continuity 1.4
Balls

Answer 23

for every ε > 0 there exists a δ > 0 such that f(Bδ(x)) ⊂ Bε(f(x)).

The map f is continuous, if it is continuous at all points of X.

or if given any x_n converging to x, f(x_n) converges to f(x) (sequential continuity the 1.41)

Question 24

THM. 1.41 SEQUENTIAL CONTINUITY

Answer 24

For f : (X,dX) → (Y,dY), f is continuous at a if and only if,

whenever a sequence x_n → a, then f(x_n) → f(a).

In short, f is continuous at a IFF f permutes with the limit:
f(lim_n→∞ ( x) )= lim_n→∞[ f(x_n )]
for any sequence x_n → a.

Question 25

thm 1.41 proof: sequential continuity

Answer 25

if f is continuous at a and x_n→ a, then for each ε > 0 we have a δ > 0 such that d_Y (f(x),f(a)) < ε whenever d_X(x,a) < δ. Then there’s an n_0 with d(x_n,a) < δ for all n ≥ n_0, and so d(f(xn),f(a)) < ε for all n ≥ n0. Thus f(xn) → f(x). Conversely, if f is not continuous at a, then there is an ε for which no δ will do, so we can find x_n with d(x_n,a) < 1/n but d(f(x_n),f(a))≥ε. Then xn→a but f(xn) does not converge to f(a).

Question 26

alt way to define continuity with preimage use in the 1.42

Answer 26

For a mapping f:X→Y and a set U⊂Y, let f^{−1}(U) be the set ={x in X: f(x) in U} This makes sense even if f−1 is not defined as a function.

Question 27

THM 1.42 continuity and open sets

Answer 27

A function f : X → Y is continuous if and only if f−1(U) is open in X for every open subset U ⊂ Y the inverse image of an open set is open

Question 28

THM 1.42 CONTINUITY AND OPEN SETS PROOF\. preimage

Answer 28

Suppose f continuous , ⊂ Y is open, and that x0 ∈ f−1(U), so f(x0) ∈ U. There is a ball B_ε(f(x0)) ⊂ U, since U is open, and then by continuity there is a δ > 0 such that d_Y (f(x),f(x0)) < ε whenever dX(x,x0) < δ. This means that for d(x,x0) < δ, f(x) ∈ U and so x ∈ f−1(U). That is, f−1(U) is open CONVERSE if the inverse image of an open set is open, and x0 ∈ X, let ε > 0 be given. We know that B_ε(f(x0)) is open, so f−1(B(f(x0),ε)) is open, and contains x0. So it contains some B_δ(x_0) with δ > 0. But now if d(x,x0) < δ, we have x ∈ Bδ(x0) ⊂ f^{−1}(Bε(f(x_0))) so f(x) ∈ Bε(f(x0)) and we have d(f(x),f(x0)) < ε.

Question 29

continuity and open sets: f : R → R, f ≡ 0, then f(R) =?

Answer 29

={0} not open- so for f continuous not always expect f(U) to be open for all open subsets of x (its the preimage of open subset must be open if continuous not the other way)

Question 30

EXAMPLE 1.44 Let X = R with the discrete metric, and Y any metric space. Then all functions f : X → Y are continuous!

Answer 30

* Because the inverse image of an open set is an open set, since all sets are open. (i) In discrete metric space (X, d) every subset of X is open. In particular, every singleton {x} is open * Because whenever xn → x0 we have xn = x0 for n large, so obviously f(xn) → f(x0).

Question 31

Exercise 1.45. Which functions from a metric space X to the discrete metric space Y = R are continuous?

Answer 31

Using thm 1.42 we require the preimage of open sets to be open, not all subsets of Y are open? Propn 1.46...

Question 32

PROP 1.46 CONTINUITY for composition and closed sets

Answer 32

Let X and Y be metric spaces. 1. A function f : X → Y is continuous if and only if f−1(F) is closed whenever F is a closed subset of Y . 2. If f : X → Y and g : Y → Z are continuous, then so is the composition g ◦ f : X → Z defined by (g ◦ f)(x) = g(f(x))

Question 33

proof continuity using closed sets and composites

Answer 33

PROOF C1. We can do this by complements, as if F is closed, then U = F^c is open, and f^{−1}(F) = f^{−1}(U)^c (a point is mapped into F if and only if it isn’t mapped into U). Then f^{−1}(F) is always closed when F is closed ⇐⇒ f^{−1}(U) is always open when U is open. 2. Take U ⊂ Z open; then (g′f)^{−1}(U) = f^{−1}(g^{−1}(U)); for these are the points which map under f into g^{−1}(U) so that they map under g′f into U. Now g^{−1}(U) is open in Y, as g is continuous, and then f^{−1}(g^{−1}(U)) is open in X since f is continuous.

Question 34

DEF 1.47 UNIFORM CONTINUITY

Answer 34

A function f : (X, dX ) → (Y, dY ) is called uni- formly continuous if for each ε > 0 there exists δ_ε > 0 such that whenever x, x′ ∈ X satisfy d_X(x,x′) ≤ δ_ε, we have that dY (f(y),f(y′)) ≤ ε. Here we have the same δ_ε for all x any UC function is continuous at every point

Question 35

f(x)=1/x continuity on (0,1) uniform?

Answer 35

not uniformly continuous but is continuous for x closer to 0 δ_ε chosen smaller than if x not close to 0 (sketch) assume it were: then given. ε=1/n there is a δ_ε>0 s.t. |(1/x)-(1/y)| = |(x-y)/xy| < δ/xy < ε =1/n for all x in (0,1) s.t |x-y|< δ, thus δ <|xy|/n for all x,y in (0,1)... continue by choosing a y in terms of δ which contradicts δ>0

Question 36

Example: The function f:R→R f(x) = {1 x <2 { 2 x ≥ 2 is it continuous? show

Answer 36

sketch: shows discontinuity at x=2 By def of continuity:for every ε > 0 there exists a δ > 0 such that f(Bδ(x)) ⊂ Bε(f(x)). there must be no δ>0 st f( B_δ(2)) ⊂ Bε(f(2)). ie choose ε=0.5 suppose d(x,2) < δ for some δ>0 and d(f(x),f(2)) < ε=0.5 for these such x. But choose x = 2- (δ/2) f(x) = 1 not possible: OR use sequential continuity e.g (x_n) = 2- 1/n which converges to 2 as n tends to infinity Does the limit commute? f(lim x_n) = f(2)=2 but lim f(2-1/n) = lim 1 = 1 so not continuous

Question 37

Example let f: ( R^2 , d_2)→R f(x,y) = x^2-y^3 show its continuous

Answer 37

To show its continuous we will show that: for every ε > 0 there exists a δ > 0 such that f(Bδ(x)) ⊂ Bε(f(x)). or For each ε>0 we will show there is a δ_ε,x >0 s.t whenever d_2((x,y),(x',y')) < δ_ε,x we have d(f(x,y), f(x',y')) But we will use sequential continuity: and using proposition 1.37 if we have (x_n,y_n) → (alb) we have x_n→a and y_n→b . This lim (f(a_n,b_n)= lim (a^2-b^3) by the algebra of limits = a^2 -b^3 =f(alb) = f(lim(x_n,y_n))

Question 38

Exercises uses subsequences:

Answer 38

If a subsequence converges then the whole sequence converges. If there is a compact metric space then...