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 .