Finally, using the continuity of both
and
applied to the positive numbers
and
choose
with
and such that if
and
then
and
Then, if
and
we have that
as desired.
- Prove part (2) of the preceding theorem.
(It's an
argument.)
- Prove part (3) of the preceding theorem.
(It's similar to the proof of part (5) only easier.)
- Prove part (4) of the preceding theorem.
- Prove part (6) of the preceding theorem.
- Suppose
is a subset of
Verify
the above theoremreplacing “ continuity”
with left continuity and right continuity.
- If
is a subset of
show that
is continuous at a point
if and only if it is
both right continuous and left continuous at
The composition of continuous functions is continuous.
Let
and
be subsets of
and let
and
be functions.
Suppose
is continuous at a point
and that
is continuous at the point
Then the composition
is continuous at
Let
be given. Because
is continuous at
the point
there exists an
such that
if
Now, using this positive number
and using the fact
that
is continuous at the point
there
exists a
so that
if
Therefore, if
then
and hence
which completes the proof.
- If
is the function defined by
prove that
is continuous at each point of
- Use part (a) and Theorem 3.2 to conclude that every rational function is continuous
on its domain.
- Prove that a step function
is continuous everywhere on
except possibly at the points of the partition
that determines
- Let
be the set of nonnegative real numbers, and define
by
Prove that
is continuous
at each point of
HINT: For
use
For
use the identity
- If
is the function defined by
show that
is continuous at every point of its domain.
Using the previous theorems and exercises, explain why
the following functions
are continuous on their domains.
Describe the domains as well.
-
-
-
- If
and
are real numbers, show that
- If
and
are functions from
into
show that
- If
and
are real-valued functions that are
both continuous at a point
show that
and
are both continuous at
Let
be the set of natural numbers, let
be the set of positive real numbers, and define
by
Prove that
is continuous at each point of
Show in fact that every function
is continuous on this domain
HINT: Show that for any
the choice of
will work.
- Negate the statement:
“For every
- Negate the statement:
“For every
there exists an
for which
- Negate the statement that “
is continuous at
The next result establishes an equivalence
between the basic
definition of continuity
and a sequential formulation.In many cases, maybe most, this sequential version of continuity
is easier to work with than the
version.
Let
be a complex-valued function on
and let
be a point in
Then
is continuous at
if and only if the following condition holds: For every sequence
of elements of
that converges to
the sequence
converges to
Or, said a different way, if
converges to
then
converges to
And, said yet a third (somewhat less precise) way, the function
converts convergent sequences to convergent sequences.
Suppose first that
is continuous at
and let
be a sequence of elements of
that converges to
Let
be given.
We must find a natural number
such that if
then
First, choose
so that
whenever
and
Now, choose
so that
whenever
Then if
we have that
whence
This shows that the sequence
converges to
as desired.
We prove the converse by proving the contrapositive statement;
i.e., we will show that if
is not continuous at
then there does exist a sequence
that converges to
but for which
the sequence
does not converge to
Thus, suppose
is
not continuous at
Then there exists an
such that for every
there is a
such that
but
To obtain a sequence, we apply this statement to
's of the form
Hence, for every natural number
there exists a point
such that
but
Clearly, the sequence
converges to
since
On the other hand, the sequence
cannot be converging to
because
is always
This completes the proof of the theorem.