Definition. If X is a set, a metric on X is a function
such that:
(a) for all
;
if and only if
.
(b) for all
.
(c) ( Triangle Inequality) For all ,
Lemma. Let X be a set with a metric, and consider the set of open balls of the form
The set for all
and all
forms a basis for a topology on X.
Proof. If , then
.
Suppose and
are open balls. Let
.
Let
Then . Therefore, the collection of open balls forms a
basis.
Definition. If X is a set with a metric, the
metric topology on X is the topology generated
by the basis consisting of open balls , where
and
.
A metric space consists of a set X together with a metric d, where X is given the metric topology induced by d.
Remark. In generating a metric topology, it suffices to consider balls of rational radius.
Example. Let and
be elements of
, and define
This gives the standard or
Euclidean metric on .
It is clear that and that
for all
. If
and
and
, then
This is only possible if for all i, and this in turn
implies that
for all i. Therefore,
.
It is obvious that for all
.
Note that , where
is the standard
norm which gives the length of a vector. Now
, where
denotes the dot product in
. By standard
properties of the dot product,
(The inequality follows from the Schwarz
inequality .) Then
Now let and
. Then
Here is a proof of the Schwarz inequality in case you ahven't seen
it. Given
, I want to show that
; I'll show that
, and the result will follow by taking
square roots.
Set ,
, and
. I want to show that
.
If , then
, and the result is obvious. Assume then
that
. For all
,
Take . The last inequality yields
This completes the proof of the Schwarz inequality.
Thus, the standard metric on satisfies the axioms for a metric.
Obviously, the metric topology is just the standard topology.
Lemma. ( Comparison Lemma
for Metric Topologies) Let d and be metrics on X
inducing topologies
and
.
is finer than
if and only if for all
and
, there is a
such that
.
Proof. Suppose first that . Let
, and let
.
is open in
, so it's open
in
. Since the open d-balls form a basis for
, there is an open ball
such that
Conversely, suppose that for all and
, there is a
such that
. I want to show that
.
Let U be open in . I want to show that it's open in
. Let
. Since the
-balls form a basis for
, there is an
such that
By assumption, there is a such that
Therefore, .
Now is a
-open set containing x and
contained in U. Since
was arbitrary, U is open in
. Therefore,
.
The standard metric on is unbounded, in the sense that
you can find pairs of points which are arbitrarily far apart.
However, you can always replace a metric with a bounded
metric which gives the same topology.
Definition. If is a metric space
and
, then Y is bounded if there
is an
such that
Lemma. Let X be a metric space with metric d. Define
(a) is a metric.
(b) d and induce the same topology on X.
Proof. (a) Let . Since
,
, and
If , then
, so
. This shows that the first metric axiom
holds.
Since , the second metric axiom holds.
To verify the third axiom, take . Begin by noting that
if either
or
, then
or
. Therefore,
Assume that and
. Then
This verifies the third axiom, so is a metric.
(b) Observe that for ,
. The idea is to apply the Comparison
Lemma, shrinking balls if necessary to make their radii less than 1.
Let and let
.
If , then
.
If , then
Therefore, the d-topology is finer than the -topology. The other
inclusion follows by simply swapping the d's and
's.
It follows that boundedness is not a topological notion, since every subset is bounded in the standard bounded metric.
Example. The square
metric on is given by
Relative to this metric, is an n-cube centered
at x with sides of length
.
First, I'll show that is a metric. Let
.
Clearly, and
. If
, then
for all i, so
.
It's also obvious that .
If , then for each j,
Therefore,
Thus, is a metric.
Lemma. induces the same topology on
as the standard metric.
Proof. The idea of the proof is depicted below.
Note that
These inequalities may be used to get -balls contained in
d-balls and d-balls contained in
-balls; by the
Comparison Lemma, this shows that the topologies are the same.
Lemma. The square metric induces the product
topology on .
Proof. If
, then
The set on the right is open in the product topology. Since the square metric-basic sets are open in the product topology, any square metric-open set is open in the product topology.
Conversely, let
It is easy to check that sets of this form comprise a basis for the product topology.
Let , so
for all i.
Define
Then
So
It follows that U is open in the square metric topology. Since the
product topology basic sets are open in the square metric topology,
any product topology open set is open in the square metric
topology.
Lemma. Metric topologies are Hausdorff.
Proof. Let be a metric space,
and let x and y be distinct points of X. Let
. Then
and
are disjoint open sets in the metric topology which
contain x and y, respectively.
Lemma. If ,
are metric spaces, the
definition
of continuity is valid. That is, a map
is continuous
at
if and only if for every
, there is a
such that if
implies that
.
Proof. First, suppose that is continuous at
. Let
, and
consider the ball
. Since this is an open set
containing
, continuity implies that there is a
such that
Now consider the conclusion to be established. Suppose satisfies
. Then
, so
.
Therefore,
, so
.
Conversely, suppose that for every , there is a
such that if
implies that
. I want to show that f is continuous.
Let , and let V be an open set in Y containing
.
I want to find a neighborhood U of x such that
.
Since the -balls form a basis for the metric topology, I may
find an
such that
. By assumption, there is a
such that if
implies that
.
Now consider the ball . This is an open set containing
x. If
, then
.
Therefore,
, so
. This shows that
, so f is continuous.
Definition. If X is a set, a sequence in X is a function .
It's customary to write for
in this situation, and to abuse
terminology by referring to the collection
as "the
sequence".
Definition. Let X be a topological space. A
sequence converges to a point
if for every neighborhood U of x, there is an integer
N such that
for all
.
means that
converges to x.
Lemma. Let X be a Hausdorff space. Convergent sequences converge to unique points.
Proof. Let and
. I want to show that
.
Suppose . Since X is Hausdorff, I can find disjoint
neighborhoods U of x and V of y. Since
, I can find an
integer M such that
implies
. Since
, I can find an integer N such that
implies
. Therefore, for
, I have
. This is
nonsense, so
.
In particular, limits of sequences are unique in metric spaces.
Lemma. ( The Sequence
Lemma) Let X be a topological space, let , and let
. If there is a sequence
with
for all n and
, then
. The converse is true if X is a
metric space.
Proof. Suppose that there is a sequence with
for all n and
. Let U be a neighborhood of x. Find an integer N
such that
for all
. Obviously, U meets Y.
This proves that
.
Conversely, suppose that X is a metric space and . For each
, the ball
meets Y, so I may choose
. I claim that
.
Let U be a neighborhood of x. Since the open balls form a basis for
the metric topology, I may find such that
; then I may find
such that
, so
.
For all , I have
, so
. Since
, I have
for all
.
This proves that .
Theorem. Let X be a metric space, let Y be a
topological space, and let . f is continuous if and only
if
in X implies that
in Y.
More succinctly, continuous functions carry convergent sequences to convergent sequences.
Proof. Suppose f is continuous, and suppose
in X. Let V be a neighborhood of
in Y. By continuity, there is a neighborhood U of x such that
.
Since , there is an integer N such that
for all
. Then
for all
. This proves that
.
Conversely, suppose that in X implies that
in Y. To show f is continuous, it will suffice to
show that for all
, I have
.
Thus, take . I want to show that
.
Now X is a metric space and , so by the Sequence
Lemma, there is a sequence of points
with
. By hypothesis, this implies that
. Since
is a sequence in
,
the Sequence Lemma implies that
. Therefore, f is
continuous.
Definition. Let be a
sequence of functions from X to Y, where Y is a metric space.
converges uniformly to a
function
if for every
, there is an
integer N such that
Theorem. Let be a
sequence of continuous functions from X to Y, where Y is a metric
space. If
converges uniformly to
, then f is
continuous.
This is often expressed by saying that a uniform limit of continuous functions is continuous.
Proof. Let and let
be a neighborhood of
. I want to find a
neighborhood U of a such that
.
First, uniform continuity implies that there is an integer N such that
In particular,
is continuous, so there is a neighborhood U of a such
that
.
Thus,
Moreover, restricting (*) to and
, I have
Therefore, the triangle inequality implies that
for all .
This proves that f is continuous.
Example. For , let
be defined by
For fixed x, .
Thus, this sequence of functions converges pointwise to the constant
function 0.
The picture below shows the graphs of for
on the interval
.
I will show that the convergence is uniform on the interval . Thus, choose
; I must find
an integer N such that if
, then
Since ,
is
an increasing function;
, so it follows
that
Now choose N such that . Then if
,
This proves that converges uniformly to 0 on
.
Example. For , let
be defined by
For fixed ,
. Thus, this sequence of functions converges
pointwise to the constant function 0 for
. It converges
pointwise to 1 for
.
The picture below shows the graphs of for
on the interval
.
I will show that the convergence is not uniform on the
interval . In fact, I will show that there is no
integer N such that if
, then
To see this, it suffices to note that , so there will always be a point in
where the function exceeds
.
Therefore, converges pointwise, but not uniformly, to
the zero function.
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