1.6) Normed And Open Closed Sets Flashcards
Defn 1.6 Normed space
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∥.
Definition 1.9. An inner product space
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).
norms on the vector space V = C[a, b]:∥f∥1
∥f∥1 =
∫ _(a,b) |f(x)| dx,
generate the respective metric d1,
norms on the vector space V = C[a, b]:∥f∥2 =
∥f∥2 =
∫ _(a,b) |f(x)|^2 dx,
generate the respective metric d2
norms on the vector space V = C[a, b]:f∥∞
∥f∥∞ = sup
x∈[a,b]
|f(x)|.
generate the respective metric d∞
norm ∥·∥2
on R^m
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
norm ∥·∥∞
on R^m
produces metrics d∞
∥·∥1: R^m →R
∥x∥∞= max{|x_1|,…,|x_n|}
Cauchy–Schwarz inequality
|⟨v, w⟩|2 ≤ ⟨v, v⟩⟨w, w⟩
in an inner product space
Cauchy–Schwarz inequality proof
start by considering the expression ⟨v + λw, v + λw⟩ ≥ 0 and analyse the
discriminant of the quadratic expression for λ.
DEFn Open set A
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”
DEFn closed set A
A ⊆ X in MS (X, d) closed if complement A^c := X\A is open
A closed set contains all its limit points
Def 1.12 open ball
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}
Def 1.12closed ball
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}.
open and closed ball relationships
x ∈ Br(x) ⊂ overline~(Br(x)) for all x ∈ X and r > 0.
Q open closed?
R open closed
Q is neither:
doesn’t contain all limit points
Complement of Q is open
R is both
Br(x) in Reals R
(x-r,x+r)
T or F: X is an open set on (X,d)
True: X on (X,d) is an open set whole metric space is open in itself.
T or F: empty set ∅ is a closed set on (X,d)
T
X on (X,d) is an open set whole metric space is open in itself so complement ∅ is considered cosed
open balls propn 1.15
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).
Def 1,22 closure of a set s̅
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)
Def 1.29 Interior
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 ∅.
Relationship open ball and int S
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)
if x is a boundary point of s
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)
X:[1,3]∪(4,∞) A= (1,3] is it open closed?
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)
(a, b) open closed
is open
[a,b] open closed
[a, b] is closed, since its complement (−∞, a) ∪ (b, ∞) is open.
[a, b) open closed
[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).
[0, 1) open? when is it
is not open in R but is open in [0,1]
X=(-∞,1]∪(3,4] C=(1,2] is C open on X
NO
There are no ɛ>0 st Bɛ(x) completely contained for any x in C
X=(-∞,1]∪(3,4] C=(1,2] is C closed on X
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
S =(-4,1) s̅?
s̅ = (-4,1)∪ {-4}∪{1} = [-4,1]
Definition 1.11 Isometry)
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
Theorem 1.18. {x_0}
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
Int s when S is open
S=int S
size of int S
less than or equal to size of S
Theorem 1.19. Collections of open subsets
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.
thm 1.19 collections of open sets proof
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).
(a,b) as a ball
(a,b)= B_{(b-a)/2} ( (b+a)/2) for a,b in reals
example set intersection or open sets not open
For the infinite set (-1/n,1/n) the intersection over N is not open
de morgans laws
A ∪ B = (B^c ∩ A^c)
(A∩B)^c = A^c ∪ B^c
example set intersection or closed sets not closes
[1/n,1] union of these over N is not closed = (0,1]
union of [-1/n,1/n] =[-1,1] is closed
Theorem 1.21 collections of closed sets
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.
Theorem 1.21 collections of closed sets proof
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
Is this set closed?
Union from n=1 to ∞)=
∪_{n=1) ^∞ [1/n,∞) = (0, ∞),
∪_{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)
Defn 1.24 Dense subset
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),
Theorem 1.25 Dense
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.
Definition 1.26 (Neighbourhood).
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)
[0,1) is it open? relate to nhds
The half-open interval [0, 1) is a neighbourhood of every point in it except
for 0.
THM 1.28 neighbourhoods and closure
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)
Propn 1.31. Interior S and closure relation
int S = X \ OVERLINE(X \ S).
The interior of S is the complement of the closure of the complement of S
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)
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)
In the metric space R we have int[0, 1) =
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)
int Q
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
PROOF prop 1.31 int S = X \ OVERLINE(X \ S).
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/
Completeness vs closedness
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.
Example: the discrete metric
Are sets open? closed?
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
