1.6) Normed And Open Closed Sets

Question 1

Defn 1.6 Normed space

Answer 1

A normed space (V, ∥·∥) is areal vector space V with a map ∥·∥ : V → R
(called norm) satisfying
1. ∥v∥ ≥ 0, and (∥v∥ = 0 ⇔ v = 0),
2. ∥λv∥ = |λ|∥v∥,
3. ∥v + w∥ ≤ ∥v∥ + ∥w∥.

Question 2

Definition 1.9. An inner product space

Answer 2

e(V,⟨·, ·⟩) is a real vector space V with a map
⟨·, ·⟩ : V × V → R (inner product) satisfying
1. ⟨λv, w⟩ = λ⟨v, w⟩,
2. ⟨v1 + v2, w⟩ = ⟨v1, w⟩ + ⟨v2, w⟩,
3. ⟨v, w⟩ = ⟨w, v⟩,
4. ⟨v, v⟩ ≥ 0, and (⟨v, v⟩ = 0 ⇔ v = 0).

Question 3

norms on the vector space V = C[a, b]:∥f∥1

Answer 3

∥f∥1 =
∫ _(a,b) |f(x)| dx,

generate the respective metric d1,

Question 4

norms on the vector space V = C[a, b]:∥f∥2 =

Answer 4

∥f∥2 =
∫ _(a,b) |f(x)|^2 dx,
generate the respective metric d2

Question 6

norms on the vector space V = C[a, b]:f∥∞

Answer 6

∥f∥∞ = sup
x∈[a,b]
|f(x)|.

generate the respective metric d∞

Question 7

norm ∥·∥2
on R^m

Answer 7

produces metric d2

∥·∥1: R^m →R
∥x∥2 = SQRT[Σ_(i=0)^(n) (x_i^2)

CAN BE WRITTEN as x.x
EUCLIDEAN NORM L^2

Question 8

norm ∥·∥∞
on R^m

Answer 8

produces metrics d∞

∥·∥1: R^m →R
∥x∥∞= max{|x_1|,…,|x_n|}

Question 9

Cauchy–Schwarz inequality

Answer 9

|⟨v, w⟩|2 ≤ ⟨v, v⟩⟨w, w⟩
in an inner product space

Question 10

Cauchy–Schwarz inequality proof

Answer 10

start by considering the expression ⟨v + λw, v + λw⟩ ≥ 0 and analyse the
discriminant of the quadratic expression for λ.

Question 11

DEFn Open set A

Answer 11

A ⊆ X in MS (X, d)
if for each point x ∈ A ∃ε > 0 st Bε(x) ⊆ A (any point has an open ball contained within the set)
(“room to swing a cat”

Question 12

DEFn closed set A

Answer 12

A ⊆ X in MS (X, d) closed if complement A^c := X\A is open

A closed set contains all its limit points

Question 13

Def 1.12 open ball

Answer 13

Let (X, d) be a metric space, let x ∈ X and
let r > 0. The open ball centred at x, with radius r, is the set
Br(x) = {y ∈ X : d(x, y) < r}

Question 14

Def 1.12closed ball

Answer 14

Let (X, d) be a metric space, let x ∈ X and
let r > 0. The closed ball centred at x, with radius r, is the set
closed ball is the set
(overline)Br(x) = {y ∈ X : d(x, y) ≤ r}.

Question 15

open and closed ball relationships

Answer 15

x ∈ Br(x) ⊂ overline~(Br(x)) for all x ∈ X and r > 0.

Question 16

Q open closed?
R open closed

Answer 16

Q is neither:
doesn’t contain all limit points

Complement of Q is open

R is both

Question 17

Br(x) in Reals R

Answer 17

(x-r,x+r)

Question 18

T or F: X is an open set on (X,d)

Answer 18

True: X on (X,d) is an open set whole metric space is open in itself.

Question 19

T or F: empty set ∅ is a closed set on (X,d)

Answer 19

T
X on (X,d) is an open set whole metric space is open in itself so complement ∅ is considered cosed

Question 20

open balls propn 1.15

Answer 20

Every “open ball” Br(x) is an open set.

proof:
For y ∈ Br(x), choose δ = r − d(x, y).
We claim that Bδ(y) ⊂ Br(x).
If z ∈ Bδ(y), i.e., d(z, y) < δ, then by the triangle inequality
d(z, x) ≤ d(z, y) + d(y, x) < δ + d(x, y) = r.
So z ∈ Br(x).

Question 21

Def 1,22 closure of a set s̅

Answer 21

The closure of S, s̅, is the smallest closed
set containing S,

and is contained in all other closed sets containing S.
s̅ = ∩ {F: F⊃S}
S⊆ s̅ and contains all limiting points for seq in A (superset of A taking all accumulation points)

Question 22

Def 1.29 Interior

Answer 22

The interior of S, int S, is the largest open set contained in
S,
int S =
[
{U : U ⊂ S, U open}.

the union of all open sets contained in S. There is at least one, namely ∅.

Question 23

Relationship open ball and int S

Answer 23

If some ball B_r(x) is contained in S then x∈ INT S ⊂ S
as int S is the union of all open balls contained in S

If there is some ball which contains no points of S then x∉S and x∈int(S^c)

Question 24

if x is a boundary point of s

Answer 24

Then every ball centred on x has both points of S and of X\S
(it will be in either of the sets but not in both)

Question 25

X:[1,3]∪(4,∞) A= (1,3] is it open closed?

Answer 25

A is open in X: Balls always contained in A
Let x∈A.
If x∈(1,3) let ɛ = min{|3-x|,|1-x|}
Bɛ (x)⊂X
If x=3 then for example B1(3)=(2,3] ⊂X (on X)

Question 26

(a, b) open closed

Answer 26

is open

Question 27

[a,b] open closed

Answer 27

[a, b] is closed, since its complement (−∞, a) ∪ (b, ∞) is open.

Question 28

[a, b) open closed

Answer 28

[a, b) is not open, since there is no open ball B(a, r) contained in the set. Nor it is closed,
since its complement (−∞, a)∪[b, ∞) isn’t open (no ball centred at b can be contained in the
set).

Question 29

[0, 1) open? when is it

Answer 29

is not open in R but is open in [0,1]

Question 30

X=(-∞,1]∪(3,4] C=(1,2] is C open on X

Answer 30

NO
There are no ɛ>0 st Bɛ(x) completely contained for any x in C

Question 31

X=(-∞,1]∪(3,4] C=(1,2] is C closed on X

Answer 31

yes
let x ∈C^c =A
so x∈ (-∞,1]∪(3,4]
if x<1 then x∈ (-∞,1]
let ɛ=0.5 then Bɛ(x) ⊆ (-∞,1] ⊆A
if x>=1 then x∈ (3,4] let ɛ=0.5|3-x| then Bɛ(x) ⊆ (-3,4] ⊆A

Question 32

S =(-4,1) s̅?

Answer 32

s̅ = (-4,1)∪ {-4}∪{1} = [-4,1]

Question 33

Definition 1.11 Isometry)

Answer 33

The class of maps, which preserve metrics:

Let (X, d_X) and (Y, d_Y ) be two metric spaces. A map
ϕ : X → Y is an isometry if

d_Y (ϕ(x_1), ϕ(x_2)) = d_X(x_1, x_2) for all x_1, x_2 ∈ X.

A metric space (X, d_X) is isometric to a metric space (Y, d_Y ) if there is an isometry
bijection between X and Y

Question 34

Theorem 1.18. {x_0}

Answer 34

In a metric space, every one-point set {x_0} is closed

proof: show that the complement, set U = {x ∈ X : x ̸= x_0} is open, so take a point x ∈ U.
Now d(x, x_0) > 0, and the ball Br(x) is contained in U for every 0 < r < d(x, x_0). x ≠x_0

Question 35

Int s when S is open

Answer 35

S=int S

Question 36

size of int S

Answer 36

less than or equal to size of S

Question 37

Theorem 1.19. Collections of open subsets

Answer 37

Let (U_α)_α∈A be any collection of open subsets of a metric space (X, d)
(not necessarily finite!).

Then
∪_{α∈A} Uα is open.

Let U and V be open subsets of a MS (X, d).
Then U ∩ V is open.

Hence (by induction) any finite intersection of open subsets is open.

Question 38

thm 1.19 collections of open sets proof

Answer 38

Proof. If x ∈(U_α)_α∈A then there is an α with x ∈ Uα.
Now Uα is open, so Br(x) ⊂ Uα for
some r > 0. Then Br(x) ⊂
(U_α)_α∈A so the union is open.

If now U and V are open and x ∈ U ∩ V , then ∃r > 0 and s > 0 such that Br(x) ⊂ U and
B(x, s) ⊂ V , since U and V are open. Then B(x, t) ⊂ U ∩ V if t ≤ min(r, s).

Question 39

(a,b) as a ball

Answer 39

(a,b)= B_{(b-a)/2} ( (b+a)/2) for a,b in reals

Question 40

example set intersection or open sets not open

Answer 40

For the infinite set (-1/n,1/n) the intersection over N is not open

Question 41

de morgans laws

Answer 41

A ∪ B = (B^c ∩ A^c)

(A∩B)^c = A^c ∪ B^c

Question 42

example set intersection or closed sets not closes

Answer 42

[1/n,1] union of these over N is not closed = (0,1]

union of [-1/n,1/n] =[-1,1] is closed

Question 43

Theorem 1.21 collections of closed sets

Answer 43

Let (F_α)α∈A be any collection of closed subsets of a metric space (X, d)
(not necessarily finite!).
Then T
∩{α∈A} Fα is closed.

Let F and G be closed subsets of a
metric space (X, d). Then F ∪ G is closed. Hence (by induction) any finite union of
closed subsets is closed.

Question 44

Theorem 1.21 collections of closed sets proof

Answer 44

Proof. uses de Morgan’s laws.
x ∉ ∪α[A_α ] means x∉A α for all α so x∈A^c_ α for all α
x ∉ ∩α[A_α ] means x∉∩A α for some α

de Morgan’s laws:
(∪A_α )^c = ∩[A^c_ α]
(∩A_α)^c = ∪[A^c_ α]

Write U_α = F^c_α = X \ F_α which is open.

So ∪α[A_n ] is open by Thm 1.19. Now, by
de Morgan’s laws, (∩{α∈A}[F_α ])^c = ∪{α∈A}(F^c α) = ∪{α∈A}[Uα]. Since its complement is
open, ∩{α∈A}[F_α] is closed.

Similarly, the complement of F ∪ G is F^c ∩ G^c, which is the intersection of two open sets
and hence open by Theorem 1.19. Hence F ∪ G is closed

Question 45

Is this set closed?
Union from n=1 to ∞)=
∪_{n=1) ^∞ [1/n,∞) = (0, ∞),

Answer 45

∪_{n=1) ^∞ [1/n,∞) = (0, ∞),
which is open but not closed

Infinite unions of closed sets do not need to be closed
thm 1.21 finite unions of closed subsets are closed ( infinite intersections of closed sets are closed)

Question 46

Defn 1.24 Dense subset

Answer 46

A subset S ⊂ X is dense in X if s̅ = X.

Ie any nhd of points in S is contained in X?

dense in X if every point of X either belongs to A or else is arbitrarily “close” to a member of A —

An alternative definition of dense set in the case of metric spaces is the following. given by a metric, the closure overline {A}} of A in X is the union of A and the set of all limits of sequences of elements in
A (its limit points),

Question 47

Theorem 1.25 Dense

Answer 47

The set Q of rationals is dense in R, with the usual metric

Q is a decimal approx for R

PROOF:
Suppose that F is a closed subset of R
F⊆R which contains Q: we claim that F = R.
For U = R \F is open and contains no points of Q. But an open set U (unless it is empty)
must contain an interval Br(x) for some x ∈ U, and hence a rational number.
Our only conclusion is that U = ∅ and F = R, so that closure (overline Q) = R.

Question 48

Definition 1.26 (Neighbourhood).

Answer 48

V is a nbh of x if there
is an open set U st x ∈ U ⊆ V ; this means that ∃δ > 0 s.t. B_δ(x) ⊆ V .

Thus a set is open precisely when it is a neighbourhood of each of its points.

(ball centred on X inside nhd V)

Question 49

[0,1) is it open? relate to nhds

Answer 49

The half-open interval [0, 1) is a neighbourhood of every point in it except
for 0.

Question 50

THM 1.28 neighbourhoods and closure

Answer 50

For a subset S of a metric space X, we have x ∈ s̅ iff V ∩ S ̸= ∅ for
all nhds V of x (i.e., all neighbourhoods of x meet S)

Question 51

Propn 1.31. Interior S and closure relation

Answer 51

int S = X \ OVERLINE(X \ S).

The interior of S is the complement of the closure of the complement of S

Question 52

Theorem 1.28.PROOF all nhds of x meet s (x is in closure of s iff nhd and intersection of S is never empty)

Answer 52

Proof.
Suppose such a nhd exists that doesnt meet s: then
V ∩ S = Ø for some nhd V of x
then there is an open subset
U with x ∈ U and U ∩ S = ∅.
Then X \ U is a closed set containing S ( as x∈ U) so S ⊂ X \ U, (x∉ X\U) and then x ∉ s̅ (Diagram: S contained in s̅ , ball centred on point in s̅ has some part in S)

Conversely, if every neighbourhood of x does meet S, then x ∈ s̅, as otherwise X \ s̅ is as
open neighbourhood of x that doesn’t meet S. If there are no points of S in nhd there is none in the ball. (diagram: ball in interior of S, completely. Ball on boundary of S in s̅ meets S)

Question 53

In the metric space R we have int[0, 1) =

Answer 53

In the metric space R we have int[0, 1) = (0, 1); clearly this is open
and there is no larger open set contained in [0, 1)

Question 54

int Q

Answer 54

int Q = ∅. For any non-empty open set must contain an interval Br(x) and then it
contains an irrational number, so isn’t contained in Q

Question 55

PROOF prop 1.31 int S = X \ OVERLINE(X \ S).

Answer 55

de Morgans:
int s = ∪ {U: U⊂ S, U is open}
= X\ (∩[{U^c: U ⊂ S, U open}])
= X\ (∩[{F: X\S ⊂ F, F is closed}])
= X(overline (X\S))

F= U^c is closed when U is open. open sets contained in s and closed sets containing its complement/

Question 56

Completeness vs closedness

Answer 56

A metric space is complete if every Cauchy sequence converges (to a point already in the space). A subset F
of a metric space X
is closed if Fcontains all of its limit points; this can be characterized by saying that if a sequence in F
converges to a point x in X

, then x must be in F. It also makes sense to ask whether a subset of X is complete, because every subset of a metric space is a metric space with the restricted metric.

It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences. However, closed subspaces need not be complete. For a trivial example, start with any incomplete metric space, like the rational numbers Q
with the usual absolute value distance. Like every metric space, Q
is closed in itself, so there you have a subset that is closed but not complete. If taking the whole space seems like cheating, just take the rationals in [0,1], which will be closed in Q but not complete.

If X is a complete metric space, then a subset of X is closed if and only if it is complete.

Question 57

Example: the discrete metric
Are sets open? closed?

Answer 57

Each single point set is open as each one is the ball radius 1/2 centred at x, thus every set is open . Since for x in U we have the ball contained in U

Ever set is also closed by taking complements