Q1
Theorem 4.19: Let
continuous on
be a continuous mapping of a compact metric space
. Then
is uniformly
.
Theorem 2.37: If
Suppose
into a metric space
is a infinite subset of a compact space
is not uniformly continuous, then exists
.
, then
has a limit point in
, sequences
means for all
, exists
in
.
such that
such that for all
, so we can take out the absolute sign to show all
, exists
but
,
. Since
such that for all
,
.
By Theorem 3.6, therefore exists subsequence of
such that exists
such that
for all
and
and
and
, denoted as
, and subsequence of
. Therefore, for all
for all
. Since
, denoted as
, exists
such that
. Therefore, by the above paragraph, we know that
is a subsequence of
. Therefore,
, therefore
.Consider
by triangle inequlity, therefore we can know that
, implying that
.
Consider whether
is a limit point of
Claim:
is a limit point of
Justication: Let
.
, if
. Consider
.
. Since
and
, exists
. Therefore,
Consider the case of
. Hence,
. Therefore,
. In other words, exists
. Let
s.t.
is a limit point of
.
. By Theorem 4.2,
,
such that
,
. Therefore, for all
Therefore, we have
, we have:
. Therefore, for all
. However, since
. Let
, exists
is a subsequence of
such that
,
, which contradicts.
Consider the case for some ,
such that
. We can extract all terms which equal to
. Therefore,
sequence that equal to . By the same reasoning above, since
,
, which contradicts.
to form 2 new subsequence
will be a constant
is a subsequence of
Q2
Consider
So
by
. Consider
, which is a closed set. Therefore,
is continuous.
is continuous at every irrational point.
Let
where
and
are integers without any common divisors. Consider
be a irrational number such that
. Therefore,
. Let
. Let
. Hence,
. Therefore, I manged to show for all rational point ,
such that
.
Q3
Suppose
and
. Claim that
. Let
such that
. Let
Therefore,
Suppose exists
Therefore,
Therefore,
. Hence,
. Also, since
. Therefore,
converges uniformly to
on
such that for all ,
on
. Let
, we have:
.
. Since
, exists
such that
, we have:
.
,
. Therefore,
converges uniformly to
, exists
. Since
.
Q4
Define
Hence
as
. Hence
uniformly converges to
uniformly converges to
. However,
. Define
as
doesn't converge uniformly to
.
but
pointwisely.
Q5
Consider
converges uniformly. Let
. Given that
such that
. Since
Q6
converges, by Theorem 7.10,
. Consider the partial sum
is continuous at
such that
for all , by Theorem 7.12,
is continuous for every
.
Consider the partial sum
such as
. Fix . Fix . Therefore,
as a geometric series that converges, by Theorem 7.10,
summation of the cosine functions,
continuous on all of
.
converges uniformly to
) is continuous for every
. Since
. Since for all ,
. Therefore, by Theorem 7.12,
as a
is