Solutions to Problem set 4

Chapter 2: Problem 10

This is the discrete metric' on a set. Certainly is well defined and iff Symmetry, is immediate from the definition and the triangle inequality

 (10)

follows from the fact that the right hand side is always equal to or and the LHS is 0 or and if the LHS vanishes then and the RHS also vanishes.

All subsets are open, since if and then and implies and hence Since the complements of open sets are closed it follows that all subsets are closed. The only compact subsets are finite. Indeed if is compact then the open balls of radius with centers in cover and each contains only one point of so the existence of a finite subcover implies that itself is finite.

Chapter 2: Problem 12

We are to show that is compact as a subset of directly from the definition of compactness. So, let be an open cover of It follows that for some But since is open it contains some ball of radius around Thus all the points for For each we can find some such that since the cover Thus we have found a finite subcover

 (11)

and if follows that is compact.

Chapter 2: Problem 16

Here is the metric space, with the usual' metric. Set

 (12)

Suppose is a limit point of as a subset of the rationals. Then we know that is infinite for each Regarding as a real number it follows that Since we know the end points are not rational and by assumption it follows that Thus is closed. Certainly is bounded since implies

To see that is not compact, recall that if it were compact as a subset of it would be compact as a subset of by Theorem 2.33. Since it is not closed as a subset of it cannot be compact. Alternatively, for a direct proof of non-compactness, take the open cover given by the open sets This can have no finite subcover since contains points arbitrarily close to the real point

Yes is open in since it is of the form where is open, so Theorem 2.30 applies.

Chapter 2: Problem 22

We need to show that the set of rational points, is dense in We can use the fact that is dense. Thus, given and there exists such that for each Thus, as points in

 (13)

This shows that is separable since we know that is countable.

Chapter 2: Problem 23

We are to show that a given separable metric space, has a countable base. The hint is to choose a countable dense subset and then to consider the collection, of all open subsets of of the form where and This is a countable union, over of countable sets so is countable. Now, we need to show that this is a base. So, suppose is a given open set. If then for some since it is open. Also, by the density of in there exists some with But then implies Thus It follows that

 (14)

Thus is written as a union of the elements of which is therefore an open base.

Chapter 2: Problem 25

We wish to show that a given compact metric space has a countable base. As the hint says, for each consider the balls of radius around each of the points of

 (15)

since each is in one of these balls at least. Now the compactness of implies that there is a finite subcover, that is there is a finite subset for each such that

 (16)

Now, set This is countable, being a countable union of finite sets. Now it follows that is dense in Indeed given and there exists with and from (16) a point in with Thus and it follows that is separable; from #23 it follows that has a countable open base.

Alternatively one can see directly that the form an open base.

Richard B. Melrose 2004-05-18