Proof that ( is a limit point of .

Suppose is a limit point of .
Then there is a sequence
in
such that
.
Since the absolute value function is continuous, it follows that
. Since
we know that
(and hence .) for all
. By the inequality
theorem for limits of sequences,
, i.e.
.

Proof that is a limit point of .

**Case 1:**- Suppose . Let for all . Then so , and clearly . Now , so is a limit point of .
**Case 2:**- . This case is left to you.

Now , so it follows that

and hence

This contradicts the fact that for all .

Then is continuous, and for all . If I did not put the requirement that be a limit point of in the above definition, I'd have

I certainly do not want this to be the case.

Proof: Suppose there were a continuous function on
such that
on
. Let
and
. Then and and so

and also

Hence we get the contradiction .

Proof: is continuous at , and on
. Let
be a sequence in
such that .
Since is continuous at , we have

i.e.,

so ; i.e., .

- a)
- b)
- c)
- d)
- (Here ).
- e)

Proof: Suppose has a limit at , and let be a continuous function with
, and for all
.
Let be a sequence in
such that . Then
is a sequence in
, so by continuity of ,

Hence, condition (8.33) holds with .

Conversely, suppose there is a number such that

I need to show that is continuous at . Let be a sequence in such that . I want to show that .

Let be a sequence in
such that
. (Such a sequence exists because is a limit point of
). Define a sequence in
by

Let and be precision functions for and respectively. Let

Then is a precision function for , since for all and all ,

Hence , and by assumption (8.34), it follows that . I now claim that , and in fact any precision function for is a precision function for . For all and all ,

This completes the proof.

I want to determine whether has a limit at , i.e., I want to know whether there is a number such that for every sequence in

If and then

Since is either or , we have

For each , define a sequence by

Then , and

Hence

It follows that has no limit at .

Let
. It is clear that maps points on the horizontal line to
other points on the line . I'll now look at the image of the parabola
under .

So maps the right half of the parabola into the vertical line , and maps the left half of the parabola to the line . Parabolas with get mapped to the upper half plane, and parabolas with get mapped to the lower half plane. The figure below shows some parabolas and horizontal lines and their images under .

Then from the continuity of the square root function and the composition theorem,

But does not exist, since and is not a limit point of .

Proof: Suppose that
and
exist. Let be any
sequence in

such that . Then is a sequence
in both
and
, so

By the sum theorem for limits of sequences,

Hence has a limit at , and .

The other parts of the theorem are proved similarly, and the proofs are left to you.

Now suppose
is a functiom from some set to
and is a
subset of . We say *is bounded on*
if is a bounded set, and
any bound for is called a *bound for * on . Thus a number
is a bound for on if and only if

We say

Let
for all
. Then is
bounded on
since

However, is not a bounded function, since

for all .

Let

( is the real part of the discontinuous function from example 8.35.)

I claim is bounded by . For all
,

(NOTE: is either or .) Hence if , then

To prove my claim, apply this result with and .

for all , and that equality holds if and only if . (This shows that is a bound for the function in the previous example.) HINT: Consider .

- 1) Decide whether is bounded, and if it is, find a bound for .
- 2) Decide whether is bounded on , and if it is, find a bound for on .
- 3) Decide whether has a limit at , and if it does, find .

- a)
- for all .
- b)
- for all .
- c)
- for all .
- d)
- for all .
- e)
- for all .