Exercises: Chapter 1, Section 3

1. Prove that and , show that if and only if for each .

Suppose that for each i. Let . Choose for each , a positive such that for every with , one has . Let . Then, if satisfies , then . So, .

Conversely, suppose that , , and is chosen as in the definition of . Then, for each i, if is in and satisfies , then . So .

2. Prove that is continuous at if and only if each is.

This is an immediate consequence of Problem 1-23 and the definition of continuity.

3. Prove that a linear transformation is continuous.

By Problem 1-10, there is an such that for all . Let and . Let . If satisfies , then . So T is continuous at .

4. Let .
1. Show that every straight line through contains an interval around which is in .

Let the line be . If , then the whole line is disjoint from . On the other hand, if , then the line intersects the graph of at and and nowhere else. Let . Then is continuous and . Since the only roots of are at 0 and , it follows by the intermediate value theorem that for all with . In particular, the line cannot intersect anywhere to the left of .

2. Define by if and if . For define by . Show that each is continuous at 0, but is not continuous at .

For each , is identically zero in a neighborhood of zero by part (a). So, every is clearly continuous at 0. On the other hand, cannot be continuous at because every open rectangle containing contains points of and for all those points , one has .

5. Prove that is open by considering the function with .

The function is continuous. In fact, let and . Let . If , then by Problem 1-4, one has: . This proves that is continuous.

Since , it follows that is open by Theorem 1-8.

6. If is not closed, show that there is a continuous function which is unbounded.

As suggested, choose to be a boundary point of which is not in , and let . Clearly, this is unbounded. To show it is continuous at , let and choose . Then for any with , one has . So,

where we have used Problem 1-4 in the simplification. This shows that is continuous at .

7. If is compact, prove that every continuous function takes on a maximum and a minimum value.

By Theorem 1-9, is compact, and hence is closed and bounded. Let (resp. ) be the greatest lower bound (respectively least upper bound) of . Then and are boundary points of , and hence are in since it is closed. Clearly these are the minimum and maximum values of , and they are taken on since they are in .

8. Let be an increasing function. If are distinct, show that .

One has . The function on the right is an increasing function of ; in particular, is bounded above by the quantity on the right for any . Now assume that the have been re-ordered so that they are in increasing order; let . Now add up all the inequalities with this value of ; it is an upper bound for the sum of the and the right hand side ``telescopes" and is bounded above by the difference of the two end terms which in turn is bounded above by .