1. (20points)
Consider
the function
where
and
are given the subspace topology. Show that
is
continuous.
Solution:
From the definition of continuity, we need to show that for every
and
we can find a
so that if
and
then
But,
( one continuation)
Since
(a second continuation)
Since
So
.
Choose
.
(a third continuation)
Since
now you have to
based on
.
Assume
,
then
so
. So choose
.
2.
a. (10 points)
Let
and
be two Metric Spaces with the same underlying set
.
Suppose that for every
there is an
such
that for all
,
and that for every
there is an
such
that for all
Show that
and
generate the same topology.
b .(20 points)
In the setting of a. , suppose that there is some
fixed real number
such that for all
.
Show that
and
generate the same topology by giving formulae for
given
, and
given
that produce the necessary set inclusions of a.
Hint:
Think of
as
measuring in miles and
as
measuring in kilometers. If you are within
miles of
something
then you certainly within
kilometers.
Solution:
a.
By
definition,
be open with respect to
if
for
any
we can find some
such that
.
But by assumption we can find
such
that
,
and hence
so
is also open with respect to
The
proof the other way is identical.
b.
First we use
to show that
so
or
Next use
to show that
3.
a. (10 points) State the Least Upper Bound Property for the Real Numbers and use it to prove:
b.
(20 points) Let
be
any continuous function. Show that such that
has a fixed point. That is there is
at
least one
with
Hint: What can one say about the continuous function
?
Note:
Solution:
a.
The Least Upper Bound Property states that every set of Real Numbers that has an upper bound has a least upper bound.
b.
We need to find some
such that
Suppose
there was no such
.
Let
We
know that
and
.
(
Let
Since
we know
We
also know that for
for
Suppose
Since
is
continuous there exists a
such
that
for all
contradiction the
previous statement.
Suppose
.
Again,
since
is continuous there exists a
such
that
for all
But since
for
any
there exists an
such that
.
In particular
But
,
again a contradiction.
Hence
or equivalently,
4. (20 points)
Let
be
any continuous function. Let
be the fixed point set
.
Prove that
is
closed.
Hint:
One version of the proof shows that
contains all its limit points,hence is closed. The informal, freshman
calculus version of this is:
so if
for all
then
Solution:
A proof that
contains
all its limit points.
Let
be a limit point of
, in particular for any
We need to show
Suppose
and
let
Since
is continuous choose
such that
.
Since
is a limit point of
we
can choose that
such
that
but then by the triangle inequality
A contradiction!
A proof that shows the complement of
is
open.
If
then
.
Choose any
such that
.
To shorten some of the formulae I will use metric space notation (
for
)
From the definition of
,
Since is continuous
is open and contains
.
and
We can choose
such
that
I claim
If not let
then
and
so
contradicting the definition of
But since
this
implies that
and
for
Thus
and since
is arbitrary
is
open.