M4800 Examination 1a

1. (20points)

Consider the function $\QTR{Large}{]}$ where and

are given the subspace topology. Show that $\QTR{Large}{f}$ 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, MATH

( one continuation)

Since

(a second continuation)

MATH Since

So . Choose .

(a third continuation)

Since

now you have to based on $\QTR{Large}{x}$ . Assume , then

so . So choose .

Notes

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a. (10 points)

Let and be two Metric Spaces with the same underlying set $\QTR{Large}{M}$ . 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 $\ \QTR{Large}{c>0}$ 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 $\QTR{Large}{d_{2}}$ as measuring in miles and $\QTR{Large}{d_{1}}$ as measuring in kilometers. If you are within $\QTR{Large}{n}$ miles of

something then you certainly within $\QTR{Large}{n}$ kilometers.

Solution:

By definition, be open with respect to $\QTR{Large}{d_{1}}$ if for any we can find some such that

. But by assumption we can find such that , and hence

so $\QTR{Large}{U}$ is also open with respect to The proof the other way is identical.

First we use to show that

Next use to show that

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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 $\QTR{Large}{f}$ has a fixed point. That is there is

at least one with

Hint: What can one say about the continuous function ?

Note:

Solution:

The Least Upper Bound Property states that every set of Real Numbers that has an upper bound has a least upper bound.

We need to find some such that Suppose there was no such .

Let We know that and . (

Let Since we know $\QTR{Large}{l<1.}$ We also know that for for $\QTR{Large}{x>l.}$

Suppose

Since $\QTR{Large}{g}$ is continuous there exists a such that for all contradiction the

previous statement.

Suppose .

Again, since $\QTR{Large}{g}$ 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,

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Two Versions

4. (20 points)

Let be any continuous function. Let be the fixed point set $\QTR{Large}{f}$ . 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 $\QTR{Large}{i}$ 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 $\QTR{Large}{f}$ 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 $\QTR{Large}{x}$ .

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 $\QTR{Large}{x}$ is arbitrary is open.