Basics of Metric Spaces

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1 Basics of Metric Spaces

1.1 Metric Spaces

1.1.1 Metric spaces: definition and examples

In Analysis and Calculus the definition of convergence was based on the notion of a distance between points, namely the standard distance between two real numbers is given by

d(x,y)=|x−y|.

Similarly, the distance between two points in the plane, given by

d(x,y)=d((x₁,x₂),(y₁,y₂))=

√

(x₁−y₁)²+(x₂−y₂)²

A metric space formalises this notion. This will give us the flexibility to talk about distances on function spaces, for example, or introduce other notions of distance on spaces.

Definition 1 (Metric Space) A metric space (X,d) is a set X together with a function d: X × X → ℝ that satisfies the following properties

d(x,y) ≥ 0; and d(x,y)=0 ⇐⇒ x=y (positive definite);
d(x,y)=d(y,x) (symmetric);
d(x,z) ≤ d(x,y)+d(y,z) (triangle inequality).

The function d is called the metric. The word distance will be used interchangeably with the same meaning.

Example 2

X=ℝ. The standard metric is given by d₁(x,y)=|x−y|. There are many other metrics on ℝ, for example
d(x,y)=|e^x−e^y|;

d(x,y)=
⎧
⎨
⎩
|x−y| if |x−y| ≤ 1,

1 if |x−y| ≥ 1.
Let X be any set whatsoever, then we can define the discrete metric
d(x,y) =
⎧
⎨
⎩
1 if x ≠ y,

0 if x=y.
X=ℝ^m. The standard metric is the Euclidean metric: if x=(x₁,x₂,…,x_m) and y=(y₁,y₂,…,y_m) then
d₂(x,y)= √

(x₁−y₁)²+(x₂−y₂)²+…+(x_m−y_m)²

.

This is linked to the inner-product (scalar product), x ·y=x₁ y₁+ x₂ y₂+…+x_m y_m, since it is just √(x−y).(x−y). We will study inner products more carefully later, so for the moment we won’t prove the (well-known) fact that it is indeed a metric.
Other possible metrics include
d_∞(x,y)=max{|x₁−y₁|,|x₂−y₂|,…,|x_m−y_m|}.

Another metric on ℝ^m comes from the generalisation of our first example:
d₁(x,y)=|x₁−y₁|+|x₂−y₂|+…+|x_m−y_m| .

These metrics d₁, d₂, d_∞ are all translation-invariant (i.e., d(x+z,y+z)=d(x,y)), and positively homogeneous (i.e., d(kx,ky)=|k|d(x,y)), see Ex. 8 for further discussion.
Take X=C[a,b]. Here are three metrics similar to above ones:
d₂(f,g)=

√

b

∫

a

| f(x)−g(x) |²   dx

.

Again, this is linked to the idea of an inner product, so we will delay proving that it is a metric.
d₁(f,g)=
b

∫

a

|f(x)−g(x)|   dx,

the area between two graphs
d_∞(f,g)=max{ |f(x)−g(x)|: a ≤ x ≤ b},

the maximum vertical separation between two graphs.

Example 3 On C[0,1] take f(x)=x and g(x)=x² and calculate

d₂(f,g)

⎛
⎜
⎜
⎝

∫

(x−x²)²   dx

⎞
⎟
⎟
⎠

1/2

√

1/30

d₁(f,g)

∫

|x−x²|   dx = 1/6, and

d_∞(f,g)

max

x ∈ [0,1]

|x−x²|=1/4.

Remark 4 Any subset of a metric space is again a metric space its own right, by restricting the distance function to the subset.

Example 5

The interval [a,b] with d(x,y)=|x−y| is a subspace of ℝ.
The unit circle {(x₁,x₂) ∈ ℝ²: x₁²+x₂²=1 } with d₂(x,y)=√(x₁−y₁)²+(x₂−y₂)² is a subspace of ℝ².
The space of polynomials P is a metric space with any of the metrics inherited from C[a,b] above.

Definition 6 A normed space (V,||· ||) is a real vector space V with a map ||·||: V → ℝ (called norm) satisfying

||v|| ≥ 0, and (||v||=0 ⇔ v=0),
||λ v|| = |λ| || v|| ,
||v+w|| ≤ ||v||+ ||w||.

Exercise 7 Prove that V is a metric space with metric d(v,w):=||v−w||.

Exercise 8

Write norms ||·||₁, ||·||₂, ||·||_∞ on ℝ^m which produces metrics d₁, d₂, d_∞ from Ex. 2.3.
Hint: see (11) and (9) below.

Show, that the following are norms on the vector space V=C[a,b]:

|| f ||₁

∫

|f(x)|   dx,

|| f ||₂

∫

|f(x)|²   dx,

|| f ||_∞

sup

x∈[a,b]

|f(x)|.

Furthermore, these norms generate the respective metrics d₁, d₂ and d_∞ from Ex. 2(4) as indicated in the previous exercise.

Definition 9 An inner product space(V,⟨·, ·⟩) is a real vector space V with a map ⟨·, ·⟩: V × V → ℝ (called inner product) satisfying

⟨ λ v,w ⟩ = λ⟨ v,w ⟩,
⟨ v₁ +v₂ ,w ⟩ = ⟨ v₁,w ⟩+⟨ v₂,w ⟩,
⟨ v,w ⟩ = ⟨ w,v ⟩,
⟨ v,v ⟩ ≥ 0, and (⟨ v,v ⟩ =0 ⇔ v=0).

Exercise 10

Prove that the Cauchy–Schwarz inequality | ⟨ v,w ⟩ |² ≤ ⟨ v,v ⟩ ⟨ w,w ⟩ holds.
Hint: start by considering the expression ⟨ v + λ w ,v + λ w⟩ ≥ 0 and analyse the discriminant of the quadratic expression for λ.
Then prove that V is a normed space with norm || v||:= ⟨ v,v ⟩^1/2.
Which of the above norms ||·||₁, ||·||₂, ||·||_∞ from Ex. 8 can be obtained from an inner product as described in the previous item?

There is a natural name for a class of maps, which preserve metrics:

Definition 11 (Isometry) Let (X, d_X) and (Y, d_Y) be two metric spaces. A map φ: X → Y is an isometry if

d_Y(φ(x₁), φ(x₂)) = d_X(x₁, x₂) for all x₁, x₂ ∈ 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.

1.1.2 Open and closed sets

Definition 12 (Open and closed balls) 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

B_r(x)={y ∈ X: d(x,y)<r },

and the closed ball is the set

B_r(x)

={y ∈ X: d(x,y) ≤ r }.

Trivial but useful observations are:

x ∈ B_r(x) ⊂ B_r(x) for all x∈ X and r>0, so neither ball is empty and every point is covered by all balls it centres;
B_r(x) ⊂ B_r+ε(x) for all x∈ X, r>0 and whatever small ε>0.

Note, that in ℝ with the usual metric the open ball is B_r(x)=(x−r,x+r), an open interval, and the closed ball is B_r(x)=[x−r,x+r], a closed interval.

For the d₂ metric on ℝ², the unit ball, B₁(0), is disc centred at the origin, excluding the boundary. You may like to think about what you get for other metrics on ℝ². What are balls in the discrete metric, Ex. 2.2?

Definition 13 (Open sets) A subset U of a metric space (X,d) is said to be open, if for each point x ∈ U there is an r>0 such that the open ball B_r(x) is contained in U (“room to swing a cat").

Clearly X itself is an open set, that is the whole metric space is open in itself. Also the empty set ∅ is also considered to be open in a trivial way.

Remark 14 Note that the property “be open” of a set depends on the metric space. For example if we consider the set [0,1] it is open in the metric space [0,1] with the standard metric, but not open in the set ℝ with standard metric.

Proposition 15 Every “open ball" B_r(x) is an open set.

Proof. For if y ∈ B_r(x), choose δ=r−d(x,y). We claim that B_δ(y) ⊂ B_r(x).

If z ∈ B_δ(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 ∈ B_r(x). □

Definition 16 (Closed set) A subset F of (X,d) is said to be closed, if its complement X ∖ F is open.

Note that closed does not mean “not open". In a metric space the sets ∅ and X are both open and closed. In ℝ we have:

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

Remark 17 As it can be seen from the definitions the property of a subset F to be open or closed depends from the surrounding space X. For example:

The interval [0,1) is open as a subset of the space [0,2] and is not open as a subset of ℝ (both are taken with the usual metric).
The same interval [0,1) is closed as a subset of the space (−1,1) and is not open as subset ℝ (again, both are taken with the usual metric).

Example 18 If we take the discrete metric,

d(x,y)=

⎧
⎨
⎩

1	if x ≠ y,
0	if x=y,

then each point {x}=B_1/2(x) so is an open set. Hence every set U is open, since for x ∈ U we have B_1/2(x) ⊆ U. Hence, by taking complements, every set is also closed.

Theorem 19 In a metric space, every one-point set {x₀} is closed.

Proof. We need to show that the set U={x ∈ X: x ≠ x₀} is open, so take a point x ∈ U. Now d(x,x₀)>0, and the ball B_r(x) is contained in U for every 0<r< d(x,x₀). □

Theorem 20 Let (U_α)_{α ∈ A} be any collection of open subsets of a metric space (X,d) (not necessarily finite!). Then ∪_{α ∈ A} U_α is open. Let U and V be open subsets of a metric space (X,d). Then U ∩ V is open. Hence (by induction) any finite intersection of open subsets is open.

Proof. If x ∈ ∪_{α ∈ A} U_α then there is an α with x ∈ U_α. Now U_α is open, so B_r(x) ⊂ U_α for some r>0. Then B_r(x) ⊂ ∪_{α ∈ A} U_α so the union is open.

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

Remark 21 Here we used a common property, which is helpful to remember: the minimum of a finite set of positive numbers is always positive (bigger than 0). However, the infimum of an infinite set of positive numbers can be zero, e.g. inf{1/n: n∈ ℕ}=0. Therefore, a transition from a given infinite set to a suitable finite set will be a reacquiring theme in our course, cf. compact set later in the course.

Thereafter, the collection of open sets is preserved by arbitrary unions and finite intersections.

However, an arbitrary intersection of open sets is not always open; for example (−1/n,1/n) is open for each n=1,2,3,…, but ∩_n=1^∞(−1/n,1/n)= {0}, which is not an open set.

For closed sets we swap union and intersection.

Theorem 22 Let (F_α)_{α ∈ A} be any collection of closed subsets of a metric space (X,d) (not necessarily finite!). Then ∩_{α ∈ A} F_α 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.

Proof.

To prove this we recall de Morgan’s laws. We use the notation S^c for the complement X ∖ S of a set S ⊂ X.

x ∉

∪

A_α

⇐⇒

x ∉A_α for all α, so (∪A_α)^c = ∩A_α^c.

x ∉

∩

A_α

⇐⇒

x ∉A_α for some α, so (∩A_α)^c = ∪A_α^c.

Write U_α= F_α^c =X ∖ F_α which is open. So ∪_{α ∈ A} U_α is open by Theorem 20. Now, by de Morgan’s laws, (∩_{α ∈ A} F_α)^c = ∪_{α ∈ A} F_α^c. This is just ∪_{α ∈ A} U_α. Since the complement of ∩_{α ∈ A} F_α is open, it 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 20. Hence F ∪ G is closed. □

Infinite unions of closed sets do not need to be closed. An example is

∞

∪

n=1

[

,∞)=(0,∞),

which is open but not closed in ℝ with standard metric.

Definition 23 (Closure of a set) The closure of S, written S, is the smallest closed set containing S, and is contained in all other closed sets containing S.

The above smallest closed set containing S does exist, because we can define

= ∩{F: F ⊃ S and F closed },

the intersection of all closed sets containing S. There is at least one closed set containing S, namely X itself.

Example 24 In the metric space ℝ the closure of S=[0,1) is [0,1]. This is closed, and there is nothing smaller that is closed and contains S.

Exercise 25 Give an example of an open ball B_r(x) and the respective closed ball B_r(x) with the same centre and radius in a metric space X, such that B_r(x) is not the closure of B_r(x). Note the slight discontent on our notations, which shall not mislead us in future.

Definition 26 (Dense subset) A subset S⊂ X is dense in X if S=X.

Theorem 27 The set ℚ of rationals is dense in ℝ, with the usual metric.

Proof. Suppose that F is a closed subset of ℝ which contains ℚ: we claim that it 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 B_r(x) for some x ∈ U, and hence a rational number within it.

Our only conclusion is that U=∅ and F=ℝ, so that ℚ=ℝ. □

Definition 28 (Neighbourhood) We say that V is a neighbourhood (nbh) of x if there is an open set U such that 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.

Example 29 The half-open interval [0,1) is a neighbourhood of every point in it except for 0.

Theorem 30 For a subset S of a metric space X, we have x∈ S iff V ∩ S ≠ ∅ for all nhds V of x (i.e., all neighbourhoods of x meet S).

Proof. If there is a neighbourhood of x that doesn’t meet S, then there is an open subset U with x ∈ U and U ∩ S=∅.

But then X ∖ U is a closed set containing S and so S ⊂ X ∖ U, and then x ∉ S because x ∈ U.

Conversely, if every neighbourhood of x does meet S, then x ∈ S, as otherwise X ∖ S is as open neighbourhood of x that doesn’t meet S. □

Definition 31 (Interior) The interior of S, intS, is the largest open set contained in S, and can be written as

intS = ∪{ U: U ⊂ S and U open }.

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

We see that S is open exactly when S=intS, otherwise intS is smaller.

Example 32

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

Proposition 33 intS=X ∖ (X∖ S).

Proof. By De Morgan’s laws,

intS

∪{ U: U ⊂ S and U open }

X ∖ ∩{U^c: U ⊂ S and U open }

X ∖ ∩{F: F ⊃ (X ∖ S) and F closed }

X ∖ (

X∖ S

This is because U ⊂ S if and only if U^c= (X ∖ U) ⊃ (X ∖ S). Also F=U^c is closed precisely when U is open. That is, there is a correspondence between open sets contained in S and closed sets containing its complement. □

1.1.3 Convergence and continuity

Let (x_n) be a sequence in a metric space (X,d), i.e., x₁,x₂,…. (Sometimes we may start counting at x₀.)

Definition 34 (Convergence) We say x_n → x (i.e., x_n converges to x) if d(x_n,x) → 0 as n → ∞.

In other words: x_n → x if for any ε>0 there exists N∈ℕ such that for all n>N we have d(x,x_n) < ε.

This is the usual notion of convergence if we think of points in ℝ^d with the Euclidean metric.

Theorem 35 Let (x_n) be a sequence in a metric space (X,d). Then the following are equivalent:

x_n → x;
for every open U with x ∈ U, there exists an N>0 such that (n>N) x_n ∈ U;
for every ε>0 there exists an N>0 such that (n>N) x_n ∈ B_ε(x).

Proof. 1 ⇒ 2 If x_n → x and x ∈ U, then there is a ball B_ε(x) ⊂ U, since U is open. But x_n → x so d(x_n,x) < ε for n sufficiently large, i.e., x_n ∈ U for n sufficiently large.

2 ⇒ 3 is obvious.

Finally, 3 ⇒ 1. If the 3 condition works for a given ε>0 and large n the inclusion x_n ∈ B_ε(x) implies d(x_n,x)<ε. □

Theorem 36 Let S be a subset of the metric space X. Then x ∈ S if and only if there is a sequence (x_n) of points of S with x_n → x.

Proof. If x ∈ S, then for each n we have B_1/n(x) ∩ S ≠ ∅ by Theorem 30. So choose x_n ∈ B_1/n(x) ∩ S. Clearly d(x_n,x) → 0, i.e., x_n → x.

Conversely, if x ∉S, then there is a neighbourhood U of x with U ∩ S=∅. Now no sequence in S can get into U so it cannot converge to x. □

This can also be phrased as follows, characterising closed set in terms of sequences.

Corollary 37 (Closedness under taking limits) A subset Y ⊂ X of a metric space (X,d) is closed if and only if for every sequence (x_n) in Y that is convergent in X its limit is also in Y.

Hence, the closure S is obtained from S by adding all possible limit points of sequences in S.

Example 38

Take (ℝ²,d₁), where d₁(x,y)=|x₁−y₁|+|x₂−y₂|, where x=(x₁,x₂) and y=(y₁,y₂), and consider the sequence (1/n,2n+1/n+1). We guess its limit is (0,2). To see if this is right, look at
d₁ ⎛
⎜
⎜
⎝ ⎛
⎜
⎜
⎝
1

n

,
2n+1

n+1

⎞
⎟
⎟
⎠ ,(0,2) ⎞
⎟
⎟
⎠ = ⎪
⎪
⎪
⎪
1

n

⎪
⎪
⎪
⎪ + ⎪
⎪
⎪
⎪
2n+1

n+1

−2 ⎪
⎪
⎪
⎪ =
1

n

+
1

n+1

→ 0

as n → ∞. So the limit is (0,2).
In C[0,1] let f_n(t)=tⁿ and f(t)=0 for 0 ≤ t ≤ 1. Does f_n → f, (a) in d₁, and (b) in d_∞?
(a)
d₁(f_n,f)=
1

∫

0

tⁿ   dt =
1

n+1

→ 0

as n → ∞. So f_n → f in d₁.
(b)
d_∞(f_n,f)=max{tⁿ: 0 ≤ t ≤ 1}=1 ¬→0

as n → ∞. So f_n ¬→f in d_∞.
Note: Say g_n → g pointwise on [a,b] as n → ∞ if g_n(x) → g(x) for all x ∈ [a,b]. If we define g(x)= {
0 for 0 ≤ x < 1,

1 for x=1,

then f_n → g pointwise on [0,1]. But g ∉C[0,1], as it is not continuous at 1.
Take the discrete metric
d₀(x,y)=
⎧
⎨
⎩
1 if x ≠ y,

0 if x=y.

Then x_n → x ⇐⇒ d₀(x_n,x) → 0. But since d₀(x_n,x)=0 or 1, this happens if and only if d₀(x_n,x)=0 for n sufficiently large. That is, there is an n₀ such that x_n=x for all n ≥ n₀.
All convergent sequences in this metric are eventually constant. So, for example d₀(1/n,0) ¬→0.

A result on convergence in ℝ^m.

Proposition 39 Take ℝ² with any of the metrics d₁, d₂ and d_∞. Then a sequence x_n=(a_n,b_n) converges to x=(a,b) if and only if a_n → a and b_n → b.

Proof. A useful observation is that for any x_n and x:

d₁(x_n,x) ≥ d₂(x_n,x) ≥ d_∞(x_n,x).

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 | b_n−b |<ε/2. Thus for any n > N=max(N_a, N_b):

ε >

⎪
⎪

a_n−a

⎪
⎪

b_n−b

⎪
⎪

= d₁(x_n,x) ≥ d₂(x_n,x) ≥ d_∞(x_n,x) ,

which shows the convergence in all three metrics.

To show the opposite, WLOG assume towards a contradiction that a_n ¬→a, that is, there exits ε>0 such that for any N there exists n>N such that | a_n−a |>ε. Then:

d₁(x_n,x) ≥ d₂(x_n,x) ≥ d_∞(x_n,x)= max{

⎪
⎪

a_n−a

⎪
⎪

b_n−b

⎪
⎪

a_n−a

⎪
⎪

>ε

showing the divergence in all three norms.

□

A similar result holds for ℝ^m in general.

Now let’s look at continuous functions again.

Theorem 40 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.

Proof. d_∞(f_n,f)=max{|f_n(x)−f(x)|:   a ≤ x ≤ b} → 0 as n → ∞, so, given ε>0 there is an N so that d_∞(f_n,f)<ε for n ≥ N. It follows that if n ≥ N then

d₁(f_n,f) =

∫

|f_n(x)−f(x)|   dx ≤

∫

ε   dx = ε(b−a),

so d₁(f_n,f) → 0 as n → ∞. □

Remark 41 It is also true that if d_∞(f_n,f) → 0 then f_n → f point-wise on [a,b]. The converse is false, cf. 38(2).

Now we look at continuous functions between general metric spaces.

Definition 42 (Continuity) Let f: (X,d_X) → (Y,d_Y) 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.

Another way of saying the same is that for every ε>0 there exists a δ>0 such that

f(B_δ(x)) ⊂ B_ε(f(x)).

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

Theorem 43 (Sequential continuity) For f as above, 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 if and only if f permutes with the limit:

⎛
⎜
⎜
⎝

lim

n→ ∞

x_n

⎞
⎟
⎟
⎠

lim

n→ ∞

⎛
⎝

x_n

⎞
⎠

(7)

for any sequence x_n → a.

Proof. Same proof as in real analysis, more or less. 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 n₀ with d(x_n,a)<δ for all n ≥ n₀, and so d(f(x_n),f(a))<ε for all n ≥ n₀. Thus f(x_n) → 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 x_n → a but f(x_n) ¬→f(a). □

But there is a nicer way to define continuity. For a mapping f: X → Y and a set U ⊂ Y, let f⁻¹(U) be the set, called pre-image or inverse image

f⁻¹(U)={ x ∈ X: f(x) ∈ U }.

This makes sense even if f⁻¹ is not defined as a function.

Theorem 44 (Continuity and open sets) A function f: X → Y is continuous if and only if f⁻¹(U) is open in X for every open subset U ⊂ Y. In short: the inverse image of an open set is open.

Proof. Suppose that f is continuous, that U⊂ Y is open, and that x₀ ∈ f⁻¹(U), so f(x₀) ∈ U. Now 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 d_X(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.

Conversely, 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(x₀) with δ>0.

But now if d(x,x₀)<δ, we have x ∈ B_δ(x₀) ⊂ f⁻¹(B_ε(f(x₀))) so f(x) ∈ B_ε(f(x₀)) and we have d(f(x),f(x₀))<ε. □

Remark 45 Note that for f continuous we do not expect f(U) to be open for all open subsets of X, for example f: ℝ → ℝ, f ≡ 0, then f(ℝ)={0}, not open.

Example 46 Let X=ℝ with the discrete metric, and Y any metric space. Then all functions f: X → Y are continuous! Indeed, in either way:

Because the inverse image of an open set is an open set, since all sets are open.
Because whenever x_n → x₀ we have x_n=x₀ for n large, so obviously f(x_n) → f(x₀).

Exercise 47 Which functions from a metric space X to the discrete metric space are continuous? Which function from the discrete metric space to ℝ are continuous?

Proposition 48 Let X and Y be metric spaces.

A function f : X → Y is continuous if and only if f⁻¹(F) is closed whenever F is a closed subset of Y.
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)).

Proof.

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.
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.

□

In many cases we may need a stronger notion.

Definition 49 (Uniform continuity) A function f: (X,d_X) → (Y, d_Y) is called uniformly 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 d_Y(f(x),f(x′))≤ε.

Note, that here the same δ_ε shall work for all x∈ X. Thus any uniformly continuous function is continuous at every point. On the other hand the function f(x)=1/x on (0,1) is continuous but not uniformly continuous.

1.2 Useful properties of metric spaces

Metric spaces may or may not have some useful properties which we are discussing in the following subsections: completeness and compactness.

1.2.1 Cauchy sequences and completeness

Recall that if (X,d) is a metric space, then a sequence (x_n) of elements of X converges to x∈ X if d(x_n,x) → 0, i.e., if given ε>0 there exists N such that d(x_n,x)< ε whenever n ≥ N. Thus, to show that a sequence is convergent from the definition we need to present its limit x which may not belong to the sequence (x_n). It would be convenient to deduce convergence of (x_n) just through its own properties without a reference to extraneous x. This is possible for complete metric spaces studied in this subsection.

Often we think of convergent sequences as ones where x_n and x_m are close together when n and m are large. This is almost, but not quite, the same thing in a general metric space.

Definition 50 (Cauchy Sequence) A sequence (x_n) in a metric space (X,d) is a Cauchy sequence if for any ε>0 there is an N such that d(x_n,x_m)<ε for all n, m ≥ N.

Example 51 Take x_n=1/n in ℝ with the usual metric. Now d(x_n,x_m)=|1/n−1/m|. Suppose that n and m are both at least as big as N; then d(x_n,x_m) ≤ 1/N. Hence if ε>0 and we take N>1/ε, we have d(x_n,x_m)≤ 1/N <ε whenever n and m are both ≥ N.

In fact all convergent sequences are Cauchy sequences, by the following result.

Theorem 52 Suppose that (x_n) is a convergent sequence in a metric space (X,d), i.e., there is a limit point x such that d(x_n,x) → 0. Then (x_n) is a Cauchy sequence.

Proof. Take ε>0. Then there is an N such that d(x_n,x)<ε/2 whenever n ≥ N. Now suppose both n≥ N and m ≥ N. Then

d(x_n,x_m) ≤ d(x_n,x)+d(x,x_m) = d(x_n,x)+d(x_m,x) < ε/2+ε/2=ε,

and we are done. □

Proposition 53 Every subsequence of a Cauchy sequence is a Cauchy sequence.

Proof. If (x_n) is Cauchy and (x_{n_k}) is a subsequence, then given ε>0 there is an N such that d(x_n,x_m) < ε whenever n, m ≥ N. Now there is a K such that n_k ≥ N whenever k ≥ K. So d(x_{n_k},x_{n_l})<ε whenever k, l ≥ K. □

Does every Cauchy sequence converge?

Example 54

(X,d)=ℚ, as a subspace of ℝ with the usual metric. Take x₀=2 and define x_n+1=x_n/2+1/x_n. The sequence continues 3/2, 17/12, 577/408,… and indeed the sequence converges in ℝ as x_n → x where x=x/2+1/x, i.e., x²=2. But this isn’t in ℚ.
Thus (x_n) is Cauchy in ℝ, since it converges to √2 when we think of it as a sequence in ℝ. So it is Cauchy in ℚ, but doesn’t converge to a point of ℚ.
Easier. Take (X,d)=(0,1). Then (1/n) is a Cauchy sequence in X (since it is Cauchy in ℝ, as seen above), and has no limit in X.

In each case there are “points missing from X”.

Definition 55 (Completeness) A metric space (X,d) is complete if every Cauchy sequence in X converges to a limit in X.

Theorem 56 The metric space ℝ is complete.

Remark 57 In parts of the literature ℝ is simply defined as the completion of ℚ. In this case one does not have to prove that ℝ is complete, but it is complete by construction. One then has to work a bit to show that it is also a field.

This is a result from the first year. Since its proof depends on the definition of ℝ we will not demonstrate it here.

Example 58

Open intervals in ℝ are not complete; closed intervals are complete.

What about C[a,b] with d₁, d₂ or d_∞?

Following our consideration in Ex. 38.2, define f_n in C[0,2] by

f_n(x)=

⎧
⎨
⎩

xⁿ	for 0 ≤ x ≤ 1,
1	for 1 ≤ x ≤ 2.

[DIAGRAM]

Then

d₁(f_n,f_m)

∫

|f_n(x)−f_m(x)|   dx

∫

|xⁿ−x^m|   dx

∫

(x^m−xⁿ)   dx if n ≥ m

m+1

−

n+1

≤

m+1

→ 0,

and hence (f_n) is Cauchy in (C[0,2],d₁). Does the sequence converge?

If there is an f ∈ C[0,2] with f_n → f as n → ∞, then ∫₀² |f_n(x)−f(x)|   dx → 0, so ∫₀¹ and ∫₁² both tend to zero. So f_n → f in (C[0,1],d₁), which means that f(x)=0 on [0,1] (from an example we did earlier). Likewise, f=1 on [1,2], which doesn’t give a continuous limit.

Similarly, (C[a,b],d₁) is incomplete in general. Also it is incomplete in the d₂ metric, as the same example shows (a similar calculation with squares of functions). We will see later that it is complete in the d_∞ metric.

Remark 59 Note that ℝ² is also complete with any of the metrics d₁, d₂ and d_∞; since a Cauchy/ convergent sequence (v_n)=(x_n,y_n) in ℝ² is just one in which both (x_n) and (y_n) are Cauchy/ convergent sequences in ℝ (cf. Prop. 39).

Similar arguments show that ℝ^k is also complete for k=1,2,3,…, and (with the same proof as for Corollary) all closed subsets of ℝ^k are complete.

If a metric space (X,d) is not complete one can always pass to its abstract completion in the following sense.

Proposition 60 (Abstract completion) Any metric space (X,d) is isometric to a dense subspace of a complete metric space, which is called its abstract completion if (X,d).

Proof.[Sketch of proof] We describe a metric space (X^′,d^′) in which X is isometric to a dense subset. Consider the space X′ of Cauchy sequences of X. We define an equivalence relation ∼ on X′ by

(x_n) ∼ (y_n) ⇔ d(x_n,y_n) → 0.

The set X^′ is defined to be the set of equivalence classes [(x_n)]. It has a well defined metric given by

d^′([(x_n)],[(y_n)]):=

lim

n → ∞

d(x_n,y_n).

One checks easily that this is metric and is well defined (does not depend on the chosen representative x_n of [(x_n)]). Now there is an injective map X → X^′ defined by sending x to the constant sequence (x,x,x,…). This map is an isometry. We can therefore think of (X,d) as a subset of (X^′,d^′). This subset is dense because every Cauchy sequence can be approximated by a sequence of constant sequences. So the only difficult bit in this construction is to show that (X^′,d^′) is complete. We will sketch the construction of a limit here. It turns out that it verifies completeness on a dense set.

Lemma 61 Suppose that (X,d) is a metric space and let Y ⊂ X be a dense set with the property that every Cauchy sequence in Y has a limit in X. Then (X,d) is complete.

Proof. Let (x_n) be a Cauchy sequence in X. Now replace x_n with another sequence y_n in Y such that d(x_n,y_n)<1/n. Then, by the triangle inequality, y_n is again a Cauchy sequence and converges, by assumption, to some x ∈ X. Then also x_n converges to x. □

Let us turn to the proof of completeness of X′. Suppose that (x_n) is a Cauchy sequence in X. Then, in X′ this sequence has the form ((x₁,x₁,…),(x₂,x₂,…),(x₃,x₃,…),…). This sequence has a limit, namely, (x_n) itself. □

Exercise 62 (Extension by continuity) Let (X,d) be a metric space and X₁ be a dense subset of X. Let f: X₁ → Y be a uniformly continuous function to a complete metric space (Y,d′). Show that there is a unique function f′: X → Y which satisfies two properties:

restriction of f′ to X₁ coincides with f, that is f′(x)=f(x) for all x∈ X₁;
f′ is continuous on X.

Furthermore, it can be shown that f′ is uniformly continuous on X. We will call f′ the extension of f by continuity and will often keep the same letter f to denote f′.

There are many important consequences of Ex. 62, in particular the following.

Corollary 63 All abstract completions of a metric space (X,d) are isometric, in other words, the abstract completions is unique up to isometry.

1.2.2 Compactness

Accordingly to a dictionary: compact—closely and firmly united or packed together. For a metric space a meaning of “closely and firmly united” can be defined in several different forms—through open coverings or convergent subsequences—and we will see that these interpretations are equivalent.

An open cover of a metric space (X,d) is a family of open sets (U_α)_{α ∈ I} such that

∪

α ∈ I

U_α=X.

A subcover of a cover is a subset I′ ⊂ I of the index set such that (U_α)_{α ∈ I′} is still a cover.

Definition 64 (Compactness) A metric space (X,d) is called compact if every open cover has a finite subcover.

Informally: a space is compact if any infinite open covering is excessive and can be reduced just to a finite one. An example of a compact set is [0,1] and example of non-compact—all reals or the open interval (0,1). An importance of this concept is clarified by Rem. 21.

Definition 65 (Sequential Compactness) A metric space (X,d) is called sequentially compact if every sequence (x_n)_{n ∈ ℕ} in X has a convergent subsequence.

The limit of a convergent sequence is called the accumulation point of {x_n}. It is instructive to compare the definitions of:

x is the limit of {x_n}:	∀ ε>0 ∃ N ∀ n>N: d(x, x_n) < ε;
x is an accumulation point of {x_n}:	∀ ε>0 ∀ N ∃ n>N: d(x, x_n) < ε.

Thereafter, x is not an accumulation point of {x_n} if for some ε>0 and some N for all subsequent n>N we have d(x, x_n)>ε.

Informally: a space is sequentially compact if there is no room to place infinite number of points sufficiently apart from each other to avoid their condensation to a limit. Taking the sequence x_n=n shows that the set of all reals is not sequentially compact. On the other hand, we know from previous years that bounded closed set in ℝⁿ every sequence has a convergent subsequence. Therefore, bounded closed sets in ℝⁿ are sequentially compact.

Exercise 66 What are compact sets in a discrete metric space? What are sequentially compact sets in a discrete metric space?

Lemma 67 Let (X,d) be a sequentially compact metric space. Then for every ε >0 there exist finitely many points x₁,…,x_n such that {B_ε(x_i)∣ i=1,…,n} is a cover.

Proof. Suppose this were not the case. Then there would exist an ε>0 such that for any finite number of points x₁,…,x_n the collection of balls B_ε(x_i) does not cover, i.e.

∪

i=1

B_ε(x_i) ≠ X.

Starting with n=1 and then inductively adding points that are in the complement of ∪_i=1ⁿ B_ε(x_i) we end up with an infinite sequence of points x_i such that d(x_i,x_k) ≥ ε. This sequence cannot have a Cauchy subsequence (required for convergence) in contradiction with the sequential compactness of X. □

Theorem 68 A metric space (X,d) is compact if and only if it is sequentially compact.

Proof. We show the two directions separately.

Compactness implies sequential compactness: Suppose that X is compact and let (x_i)_{i ∈ ℕ} be a sequence. We want to show that it has a convergent subsequence. Suppose (x_i) did not have a convergent subsequence. Then no point x is an accumulation point. Therefore, for each x ∈ X there exists an ε(x)>0 such that only finitely many i ∈ ℕ for which x_i ∈ B_ε(x). Since (B_ε(x))_{x ∈ X} is an open cover it has a finite subcover, that is a finite number of balls with a finite number of x_i in each. This contradicts to the infinite number of elements in the sequence (x_i).

Sequential compactness implies compactness: This implication is quite tricky. The proof is again by contradiction. Let us assume our space is sequentially compact and there exists a cover U_α that does not have a finite subcover. By the above lemma there are finitely many points x₁,…,x_N₁ such that B₁(x_i) is a cover. Each of the balls B₁(x_i) is covered by U_α as well. Since our cover does not have a finite subcover one of the balls B₁(x_i) does not have a finite subcover. Denote the relevant point x_i by z₁.

Again there are finitely many points x′₁,…,x′_N₂ such that B_1/2(x′_i) is a cover of X. The collection of sets B₁(z₁) ∩ B_1/2(x′_i), with i=1,…,N₂ is also a covering of B₁(z₁). In the same way as before there is at least one of the x′_i (which we will again call z₂), such that B₁(z₁) ∩ B_1/2(z₂) can not be covered by a finite subcover of U_α. Continuing like this we construct a sequence of points z_i such that none of the sets

B₁(z₁) ⋂ B

(z₂) ⋂ … ⋂ B

(z_N)

can be covered by a finite subcover of U_α.

By assumption the sequence (z_i) has a convergent subsequence. Say z is a limit point of that subsequence. Since U_α is an open cover the point z is contained in one of the U_α and of course that means that an open ball B_ε(z) around z is contained in U_α for some ε>0.

Now we show that there exits an N ∈ ℕ such that B_1/N(z_N) is a subset of U_α (this will be the desired contradiction!). Indeed, choose N large enough so that d(z_N,z) + 1/N<ε. Then x ∈ B_1/N(z_N) implies that d(x,z) ≤ d(z_N,z) + d(x,z_N) < d(z_N,z) + 1/N<ε. This means in particular that

B₁(z₁) ⋂ B

(z₂) ⋂ … ⋂ B

(z_N)

is a subset of U_α. Thus, there is a subcover of the set B₁(z₁) ∩ … ∩ B_1/N(z_N) consisting of one element U_α. This is a contradiction as we constructed the sequence of balls in such a way that these sets cannot be covered by a finite number of the U_α. □

Definition 69 (Boundedness) A subset A ⊂ X of a metric space is called bounded if there exists x₀ ∈ X and C>0 such that for all x ∈ A we have d(x₀,x) ≤ C.

Remark 70 One can easily see, using the triangle inequality, that the reference point x₀ can be chosen as any point in X. This means if A ⊂ X is bounded and x₀ ∈ X, then there exist a C>0 such that d(x₀,x) ≤ C for any x ∈ A.

Theorem 71 Suppose that A ⊂ X is a compact subset of a metric space. Then A is closed and bounded.

Proof. First we show A is bounded. Choose any x₀ ∈ X and note that the set B_n(x₀) indexed by n ∈ ℕ is an open cover of A. Hence, there exists a finite sub-cover< B_n₁(x₀),…,B_{n_N}(x₀). Hence, A ⊂ B_C(x₀), where C= max{n₁,…,c_N}. Hence, A is bounded.

Next assume that (x_k) is a sequence in A that converges in X. Since A is compact there exists a subsequence that converges in A. Hence, the limit of x_k must also be in A. Therefore, A is closed. □

The converse of this statement is not correct in general. It is however famously correct in ℝ^m.

Theorem 72 (Heine–Borel) A subset K ⊂ ℝ^m is compact if and only if it is closed and bounded.

Proof. We just need to combine the above statements. We have already shown that compactness implies closedness and boundedness. If K is closed and bounded we know from Analysis that it is sequentially compact. Therefore it is compact. □

As an illustration of further nice properties of compact spaces we mention the following result:

Exercise 73

Any continuous function on a compact set is bounded.
Any continuous function f: K→ X from a compact space K to a metric space X is uniformly continuous.

Remark 74 Note that there are two different sorts of properties of metric spaces:

the first sort of absolute properties can be verified on a metric space itself;
the second sort of relative properties is meaningful only for subsets of another metric spaces. Such a property may be true for X as a subspace of X but false if X is considered as a subspace of a different space Z.

Completeness and compactness are of the first sort, closedness is of the second, cf. Rem 17.


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Last modified: November 6, 2024.