Analysis with an introduction to proof 5th by steven lay ch03f

Since a set may have more than one metric defined on it, we often identify both and denote the metric space by X, d.. When we refer to as a metric space and do not specify any particula

Chapter 3

The Real Numbers

Section 3.6 Metric Spaces

Definition 3.6.1

Let X be any nonempty set A function d : X × X →   is called a metric on X if it satisfies the following conditions for all x, y, z ∈ X.

(1) d(x, y) ≥ 0 (2) d(x, y) = 0 if and only if x = y

(3) d(x, y) = d( y, x)

A set X together with a metric d is said to be a metric space Since a set may have more

than one metric defined on it, we often identify both and denote the metric space by (X, d ).

Example 3.6.2

(a) Let X = and define d  :  × → by d  (x, y) = |  x – y | for all x, y ∈

The fact that d is a metric follows directly from the properties of absolute values

In particular, condition (4) follows from the triangle inequality of Theorem 3.2.10(d)

When we refer to as a metric space and do not specify any particular metric,

it is understood that we are using this absolute value metric

Example 3.6.2

(b) Let X = × = and define d :  × → by

for points (x1, y1) and ( x2, y2) in 2

This metric is called the Euclidean metric on because it corresponds to our usual measure of distance between two points in the plane

2

We can now see why condition (4) is called the triangle inequality If x, y, and z are the

vertices of a triangle, then (4) states that the length of one side of the triangle must be less than or equal to the sum of the lengths of the other two sides

x

z

y

d  (x, y)≤ d (x, z) + d (z, y)

Example 3.6.2

(c) Let X be a nonempty set and define the “discrete” metric d on X by

The first three conditions of a metric follow directly from the definition of d.

The triangle inequality can be established by considering the separate cases when

the points x, y, and z are distinct or not.

If x ≠ z, then d  (x,  y) ≤ 1 ≤ 1 + d (z, y) = d  (x,  z) + d  (z,  y).

If x = z = y, then d  (x,  y) = 0 = 0 + 0 = d  (x,  z) + d  (z,  y).

If x = z but x ≠ y, then z ≠ y so that d  (x,  y) = 1 = 0 + 1 = d  (x,  z) + d  (z,  y).

From this example we see that any nonempty set can be made into a metric space.

Definition 3.6.3

Let (X,  d ) be a metric space, let x ∈ X, and let ε> 0 The neighborhood of x of radius ε

is given by

( , )

x y

d x y

=





Example 3.6.4

Let’s look again at the examples of metric spaces defined above and see what the neighborhoods look like geometrically

(a) The metric d defined on by d  (x,  y) = |  x – y | is the usual measure of distance in

(b) The Euclidean metric on produces neighborhoods that are circular disks

2

In particular, the neighborhood of

radius 1 centered at the origin 0 = (0, 0) in is given by

N(0; 1) = {(x, y) : x2 + y2 < 1}.

2

2 – 2

– 2 2

(c) The neighborhoods for the discrete metric defined

in Example 3.6.2(c) depend on the size of the radius.

If ε ≤ 1, then the neighborhood contains only the center point itself

If ε  > 1, then the neighborhood contains all of X

In particular, if X = 2 , then N  (0; 1) = {0} and N (0; 2) = . 2

N(0; 1)

To see that the triangle

It is clear that the first three conditions of a metric are satisfied by d1.

Then d1(p1, p2) = | x1 – x2 | + | y1 – y2 |

= | x1 – x3 + x3 – x2 | + | y1 – y3 + y3 – y2 |

≤ | x1 – x3 | + | x3 – x2 | + | y1 – y3 | + | y3 – y2 |

= | x1 – x3 | + | y1 – y3 | + | x3 – x2 | + | y3 – y2 |

= d1(p1, p3) + d1(p3, p2)

2 – 2

– 2 2

Geometrically, the neighborhoods in this metric are diamond shaped

N(0; 1)

If (X,  d ) is a metric space, then we can use Definition 3.6.3 for neighborhoods to

characterize interior points, boundary points, open sets, and closed sets, just as we did

in Section 3.4 for .

Most of the theorems from Section 3.4 that relate to these concepts also carry over to this more general setting with little or no change in their proofs

It follows that N  (  y; δ ) ⊆ N  (x;  ε ),

and so y is an interior point of N  (x;  ε), and N  (x;  ε ) is open ♦

One result that does require a new proof is the following theorem

Theorem 3.6.6

Let (X,  d ) be a metric space Any neighborhood of a point in X is an open set.

Proof: Let x ∈ X and let ε> 0 To see that N (x;  ε ) is an open set, we shall show that

any point y in N  (x;  ε) is an interior point of N  (x;  ε ). If y ∈ N(x  ; ε ), then

We claim that N  (  y;  δ ) ⊆ N  (x;  ε ).

N  (x;  ε ).

If z ∈ N  (  y; δ ), then d  (z,  y) < δ

It follows that d (z, x) ≤ d (z, y) + d ( y, x)

< δ + d ( y, x)

= [ε – d (x, y)] + d ( y, x) = ε

Thus z ∈ N  (x;  ε ).

The definition is the same: a set S is compact iff every open cover of S contains a finite subcover.

The definition of a deleted neighborhood in a metric space (X,  d ) is similar to that in :

The definition of an accumulation point is also analogous: x is an accumulation point

of a set S if for every Once again, the closure of a set S, ε> 0, N*(x;  ε ) ∩ S ≠ ∅

denoted by cl S, is given by cl S ∪ S′ , where S ′ is the set of all accumulation points of S.

The properties of closure and closed sets given in Theorem 3.4.17 continue to apply

in a general metric space Indeed, the proofs given there were all stated in terms of neighborhoods, so they carry over with no change at all

But compact sets are different

To see this, we first need to define what it means for a set to be

bounded in (X,  d )

Definition 3.6.10

A set S in a metric space (X,  d ) is bounded if S ⊆ N  (x;  ε ) for some x ∈ X and some ε> 0.

But the Heine-Borel theorem no longer holds

Theorem 3.6.11

Let S be a compact subset of a metric space (X,  d ) Then

(a) S is closed and bounded.

(b) Every infinite subset of S has an accumulation point in S.

entirely in terms of neighborhoods and open sets It applies without any changes

in this more general setting

except that the compactness of S is assumed (instead of using boundedness and the Heine-Borel theorem) We know that the accumulation point of S must be in S, since S is closed ♦

While a compact subset of (X,  d ) must be closed and bounded, the converse does

not hold in general The proof of the converse in (given in Theorem 3.5.5) was

very dependent on the completeness of , a property not shared by all metric spaces.

But neighborhoods are open sets by Theorem 3.6.6,

so each singleton set consisting of one point is an open set

We have seen that

To see this,

for each point p ∈ S, let Ap = {p}

But F contains no finite subcover of S,

Then S is a closed set, and it is also bounded since S ⊆ N  (0; 2) = 2

Example 3.6.12

Consider the set with the discrete metric of Example 3.6.2(c).2

Since every set is the union

of the points in the set, this means that every set is open!

Since the complement of an open set is closed, this means that every subset is also closed!

Let S be the unit square: S ={(x, y): 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1}.

We claim that S is not compact.

Then each Ap is an open set and

Thus F = {Ap : p ∈ S} is an open cover for S.

We conclude that in this metric space the unit square S is closed and bounded, but not

compact

.

S

∈