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Metric Spaces
Definition
A metric on a set X is a real-valued function d on X × X such
that, for all points x, y , z ∈ X , the following properties hold:
Functional Analysis
Lecture 1: Metric Spaces – Metrics, Open and Closed Sets
(i) d is non-negative: d(x, y ) ≥ 0;
(ii) d is definite: d(x, y ) = 0 if and only if x = y ;
(iii) d is symmetric: d(x, y ) = d(y , x);
Bengt Ove Turesson
(iv) d satisfies the triangle inequality:
d(x, z) ≤ d(x, y ) + d(y , z).
August 30, 2016
If d is a metric on X , then the pair (X , d) is called a metric space.
Remark
When the metric d on a set X is given, we will sometimes also call
X a metric space.
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Examples of Metric Spaces
Examples of Metric Spaces
Example
A metric on Rd is given by
d2 (x, y ) = |x − y | =
X
d
)t
j=1
|xj − yj |2
1/2
)t
Example
,
Rd .
where x = (x1 , . . . , xd and y = (y1 , . . . , yd are vectors in
metric on Cd . This will be the standard metric on Rd and Cd .
Other examples of metrics on Rd and Cd are
dp (x, y ) =
X
d
j=1
|xj − yj |p
1/p
Let `∞ (M) denote the set of all bounded, complex-valued
functions on an arbitrary set M.
This is also a
A metric on `∞ (M) is given by
d∞ (f , g ) = sup |f (x) − g (x)|,
,
x∈M
where f , g ∈ `∞ (M).
where 1 ≤ p < ∞, and
d∞ (x, y ) = max |xj − yj |.
1≤j≤d
Here, the triangle inequality in the case 1 ≤ p < ∞ requires a proof, which we
shall come back to later.
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Examples of Metric Spaces
Examples of Metric Spaces
Example
Example
For 1 ≤ p < ∞, the set `p consists
of all sequences x = (xj )∞
j=1 of
P
p < ∞.
complex numbers such that ∞
|x
|
j=1 j
A metric on `p is given by
(a) The metric space `∞ = `∞ (N) consists of all bounded
sequences x = (xj )∞
j=1 of complex numbers. By the definition
in the previous example,
d∞ (x, y ) = sup |xj − yj |,
j≥1
x, y ∈ `∞ .
dp (x, y ) =
X
∞
j=1
|xj − yj |p
1/p
x, y ∈ `p .
,
The triangle inequality for `p follows from Minkowski’s inequality:
∞
(b) The metric space c consists of all sequences x = (xj )∞
j=1 ∈ `
such that the limit limj→∞ xj exists.
X
∞
(c) The metric space c0 consists of all sequences
∞ such that lim
x = (xj )∞
j→∞ xj = 0.
j=1 ∈ `
j=1
(d) The metric space c00 consists of all sequences
∞ such that x = 0 for all but a finite number
x = (xj )∞
j
j=1 ∈ `
of indices j.
p
|xj + yj |
1/p
≤
X
∞
j=1
p
|xj |
1/p
+
X
∞
j=1
p
|yj |
1/p
,
p
∞
p
which holds for all sequences x = (xj )∞
j=1 ∈ ` and y = (yj )j=1 ∈ ` .
The proof of this inequality shall be given later.
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Examples of Metric Spaces
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Balls in Metric Spaces
Definition
Example
Let (X , d) be a metric space. If x ∈ X and r > 0, the open ball
Br (x) with center x and radius r is the set
(a) The space Cb (M) consists of all bounded, continuous,
complex-valued functions on a set M ⊂ Rd . The metric on
Cb (M) is the same as in `∞ (M), i.e.,
d∞ (f , g ) = sup |f (x) − g (x)|,
x∈M
Br (x) = {y ∈ X : d(x, y ) < r }.
The corresponding closed ball is B r (x) = {y ∈ X : d(x, y ) ≤ r }.
f , g ∈ Cb (M).
Example
(b) If K ⊂ Rd is compact (i.e., closed and bounded), then, as we
know from Calculus, all continuous functions on K are
bounded. We shall use the notation C (K ) for Cb (K ).
If x ∈ R and r > 0, then
Br (x) = (x − r , x + r )
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and B r (x) = [x − r , x + r ].
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Interior Points and the Interior
Open Subsets of Metric Spaces
Definition
Definition
Suppose that (X , d) is a metric space and A is a subset of X .
Suppose that (X , d) is a metric space. A subset G of X is said to be open if
all its points are interior points.
A point x ∈ A is called an interior point of A if there exists a
number r > 0 such that Br (x) ⊂ A.
Remark
Thus, G is open if G = G ◦ .
A◦
The interior
of A is the subset of A that consists of all
interior points of A.
Example
Example
An open ball Br (x) in a metric space (X , d) is of course open.
Indeed, suppose that y ∈ Br (x) and y 6= x.
Then the ball Bs (y ), where s = r − d(x, y ) > 0, is a subset of Br (x),
since if z ∈ Bs (y ), then
Let us show that [a, b]◦ = (a, b) if −∞ < a < b < ∞.
If x ∈ (a, b), then (x − r , x + r ) ⊂ [a, b], where
r = min{|x − a|, |x − a|}, showing that x ∈ [a, b]◦ .
d(x, z) ≤ d(x, y ) + d(y , z) < d(x, y ) + s = r .
Hence, (a, b) ⊂ [a, b]◦ .
This shows that y is an interior point of Br (x).
Since Br (x) ⊂ Br (x), x is also an interior point of Br (x).
Since a, b ∈
/ [a, b]◦ , it follows that [a, b]◦ = (a, b).
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Open Subsets of Metric Spaces
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Open Subsets of Metric Spaces
Proposition
Example
Suppose that (X , d) is a metric space and A is a subset of X .
Then the interior A◦ of A is an open subset of A.
(a) By the previous example, any interval of the form (a, b) ⊂ R,
where −∞ < a < b < ∞, is open: Take x as the midpoint of
the interval and r as half its length.
Proof.
This is obvious if A◦ = ∅, so suppose that A◦ 6= ∅.
(a) It is easy to see that any interval of the form (−∞, a) or
(a, ∞), where −∞ < a < ∞, is open.
Take x ∈ A◦ and choose r > 0 such that Br (x) ⊂ A. We need
to show that Br (x) ⊂ A◦ .
If y ∈ Br (x), then, since Br (x) is open, there exists a number
s > 0 such that Bs (y ) ⊂ Br (x).
Example
In any metric space (X , d), the whole space X and ∅ are open.
Since Br (x) ⊂ A, it follows that Bs (y ) ⊂ A and hence that
y ∈ A◦ .
This shows that Br (x) ⊂ A◦ .
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Open Subsets of Metric Spaces
The Topology on a Metric Space
Proposition
Remark
Suppose that (X , d) is a metric space and A is a subset of X .
Then the interior A◦ of A is the largest open subset of A with
respect to inclusion.
The following theorem shows that every metric space is what is known
as a topological space. A topological space is a set X together with a
collection τ of subsets of X that satisfies the axioms (i)–(iii) below.
The elements of τ are called open sets.
Proof.
Theorem
We already know that A◦ is an open subset of A and need to
show that any open subset B of A is also a subset of A◦ .
Suppose that (X , d) is a metric space. The collection τ of open subsets
of X is a topology on X :
We can obviously assume that B is non-empty.
(i) X , ∅ ∈ τ ;
S
(ii) if Gα ∈ τ for every α ∈TA, then α∈A Gα ∈ τ ;
n
(iii) if G1 . . . , Gn ∈ τ , then j=1 Gj ∈ τ .
If x ∈ B, then there exists a number r > 0 such that
Br (x) ⊂ B.
Since B ⊂ A, it follows that Br (x) ⊂ A and therefore that
x ∈ A◦ .
Remark
We remark that the index set A in (ii) may be infinite and even
uncountable.
Hence, B ⊂ A◦ .
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The Topology on a Metric Space
Closed Subsets of Metric Spaces
Proof.
(i) The first
S statement in the theorem is obvious.
(ii) If x ∈ α∈A Gα , then x belongs to some set Gα0 , and since Gα0
is open, Br (x)S⊂ Gα0 for some r > 0. But from this it follows
that Br (x) ⊂ α∈A Gα , which shows that the union is open.
(iii) This is left as an exercise.
Definition
Remark
(a) Any interval of the form [a, b] ⊂ R is closed since the
complement of [a, b] is the union of the sets (−∞, a) and
(b, ∞), both of which are open.
Suppose that (X , d) is a metric space. A subset F of X is said to
be closed if its complement F c is open.
Example
Notice that an infinite intersection of open sets may not be open.
For instance,
∞ \
1 1
− ,
= {0},
n n
(b) In the same manner, one can show that all intervals of the
form (−∞, a] and [a, ∞), where −∞ < a < ∞, are closed.
n=1
and {0} is not an open subset of R.
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Closed Subsets of Metric Spaces
Theorem
Suppose that (X , d) is a metric space. Then the following properties
hold:
(i) The sets X , ∅ are closed.
T
(ii) If Fα is closed for every α ∈ A,
Snthen α∈A Fα is closed.
(iii) If F1 . . . , Fn are closed, then 1 Fj=1 is closed.
Proof.
Property (i) follows from the previous theorem. We will prove that (ii)
holds and leave (iii) as an exercise. By De Morgan’s law,
\ c
[
Fα
=
Fαc .
α∈A
α∈A
Now, every set Fαc is open, so it follows that
S
c
α∈A Fα
is open.
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