1. Chapter 3
The Real Numbers

2. Section 3.6
Metric Spaces

3. 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.
d(x, y)  0
d(x, y) = 0 if and only if x = y
d(x, y) = d( y, x)
d(x, y)  d(x, z) + d(z, y) (triangle inequality)
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 3.2.9(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.

4. 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)

5. 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 = = d (x, z) + d (z, y).
If x = z but x  y, then z  y so that d (x, y) = 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 e > 0. The neighborhood of x of radius 
is given by
N (x; ) = { y  X : d (x, y) <  }.

6. Example 3.6.4
Let’s look again at the examples of metric spaces defined above and see what the
neighborhoods look like geometrically.
The metric d defined on by d (x, y) = | x – y | is the usual measure of distance in .
The neighborhoods are just open intervals: N (x; ) = (x – , x + ). 2
– 2
(b) The Euclidean metric on produces neighborhoods
that are circular disks. 2
N(0; 1)
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
(c) The neighborhoods for the discrete metric defined
in Example 3.6.2(c) depend on the size of the radius.
If e  1, then the neighborhood contains only the center point itself.
If e  > 1, then the neighborhood contains all of X.
In particular, if X = , then N (0; 1) = {0} and N (0; 2) = . 2

7. Example 3.6.5 For another example, let X = and define d1 by2
To see that the triangle
inequality also holds, let p1 = (x1, y1), p2 = (x2, y2), and p3 = (x3, y3) be arbitrary points in . 2
It is clear that the first three conditions of a metric are satisfied by d1.
Then d1(p1, p2) = | x1 – x2 | + | y1 – y2 | 2
– 2
= | x1 – x3 + x3 – x2 | + | y1 – y3 + y3 – y2 |
 | x1 – x3 | + | x3 – x2 | + | y1 – y3 | + | y3 – y2 |
N(0; 1)
= | x1 – x3 | + | y1 – y3 | + | x3 – x2 | + | y3 – y2 |
= d1(p1, p3) + d1(p3, p2)
Geometrically, the neighborhoods in this metric are diamond shaped.
If (X, d ) is a metric space, then we can use Definition 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.

8. One result that does require a new proof is the following theorem.
Let (X, d ) be a metric space. Any neighborhood of a point in X is an open set.
Proof: Let x  X and let e > 0. To see that N (x; ) is an open set, we shall show that
any point y in N (x; ) is an interior point of N (x; ).
If y  N(x ; ), then
 =   d (x, y) > 0.
N (x; ).
We claim that N ( y;  )  N (x; ).
If z  N ( y;  ), then d (z, y) < d .
It follows that d (z, x)  d (z, y) + d ( y, x)
x 
<  + d ( y, x)
y 
= [ – d (x, y)] + d ( y, x) =  .
N ( y;  )
z 
Thus z  N (x; ).
It follows that N ( y;  )  N (x; ),
and so y is an interior point of N (x; ), and N (x; ) is open. 

9. The definition of a deleted neighborhood in a metric space (X, d ) is similar to that in :
N*(x; ) = { y  X : 0 < d (x, y) <  }.
The definition of an accumulation point is also analogous: x is an accumulation point
of a set S if for every e > 0, N*(x; )  S  .
Once again, the closure of a set 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 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.
The definition is the same: a set S is compact iff
every open cover of S contains a finite subcover.
But compact sets are different.
But the Heine-Borel theorem no
longer holds.
To see this, we first need to define what it means for a set to be
bounded in (X, d ).
Definition
A set S in a metric space (X, d ) is bounded if S  N (x;  ) for some x  X and some e > 0.

10. Theorem
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.
Proof: (a) The first half of the proof of the Heine-Borel Theorem was given
entirely in terms of neighborhoods and open sets. It applies without any changes
in this more general setting.
(b) This proof is the same as the proof of the Bolzano-Weierstrass Theorem 3.5.6,
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.

11. Example
Consider the set with the discrete metric of Example 3.6.2(c). 2
We have seen that
N(p; 1) = {p} for every p  . 2
But neighborhoods are open sets by Theorem 3.6.6,
so each singleton set consisting of one point is an open set.
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!
Then S is a closed set,
and it is also bounded since S  N (0; 2) = 2
Let S be the unit square: S = {(x, y): 0  x  1 and 0  y  1}.
To see this,
for each point p  S, let Ap = {p}.
We claim that S is not compact.
Then each Ap is an open set and
But F contains no finite subcover of S,
since each set in F covers only one point in S and there are infinitely many points in S.
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.