Since we will want to consider the properties of continuous functions in settings other than the Real Line, we review the material we just covered in the more general setting of Metric Spaces.
A Metric Space,
is a set
and a function
.
such that for all
:
Positivity:
If
and
Symmetry:
Triangle Inequality:
The Triangle Inequality has a second, equivalent form:
For all
Proof:
Note that both formulas are symetric in
and
thus
we may assume that
. Thus can restate the formula in the statement
of 2.2.2 as
Adding
to or subtracting it from one inequality or the other gives the desired
equivalence.
For
where, for example,
For
more
generally, for
For
Note: For
Note:
For
the
three metrics agree.
Exercise:
For
Caculate
and
.
In fact the metrics generate the same "Topology" in a sense that will be made precise below.
For any Metric Space
is
also a metric space. The base is not important.
We will also want to understand the topology of the circle,
There are three metrics illustrated in the diagram.
Unlike the metrics for
, there is no simple function relationship between these metrics. However,
they also all generate the same "Topology" on
Fix a Metric Space
and a point
Given
we
define the open ball of radius
around
We say
is open if for any
there exist an
such
that
The set of open sets is called the Topology defined by the Metric.
For any
is open .
For every
. We need to find some
such that
Let
.
Choose
For
every
. We have
.
Thus
.
In a Metric Space
and
(the empty set) is open
Let
be an arbitrary set of open sets, then
is
also open.
Let
be
a finite set of open sets, then
is
also open.
1. and 2. follow directly from the definitions, however if you haven't been through this material you should write down the details.
3. makes use of the observation that if
then
.
Choose
. Let
. We have
for all
hence
.
Given Metric Spaces
,
and a function
.
We say that
is continuous a point
if given any
there
is a
such
that
.
We say that
is continuous if it is continuous at every point.
Exercise: Convince yourself that for
this is just the usual
definition.
Given Metric Spaces
and
, a function
is continuous if and only for every open set
is open in
Proof:
Suppose for every open set
we know that
is open in
Suppose we are given
and
.
We need to find
such that
.
But
is open. Hence
is open. Hence we can find
such that
. Or, equivalently,
Suppose
is continuous and
is open we need to show that
is open in
Suppose
we are given
and
such that
Since
is continuous, we can find
such that
Thus
Continuity is determined by the underlying Topology. That is, if two Metrics define the same open sets then functions are continuous with respect to the first metric if and only if they are continuous with respect to the second.
Suppose we have three Metric Spaces
,
, and
suppose
and
are continuous, then so is
Proof:
Let
be open. Since
is continuous so is
is open in
and since
is continuous
so is
Returning to the example of the tree metrics on the circle,
we want to show that the three Metrics
produce the same set of continuous functions.
Solution:
We need to show that
is open with respect to
iff it is open with respect to
iff
it is open with respect to
To accomplish this, it suffices to show that if we are given
and
we can find, in sequence
,
and
such
that
One note: there is no loss of generality in assuming that the
or
we work with for the next computation are smaller than the one we initially
are given, or compute.
Beginning with a simple case, let
Suppose
we know that
then we know
So
.
The hardest calculations involve
. Given
,
I need find
In computing
we make use of the inequality
for small positive values of
.
Let
then
and
for
small
Now let
. Suppose
we have
or
The computation of
follows in a similar fashion using the observation that
for
.