## Solutions to Problem set 5

Chapter 2: Problem 19

a) If and are closed in and disjoint then and so and follow from and hence the sets are separated. b) If and are open and disjoint then is closed so which means similarly so they are separated. c) With fixed and put and If is a limit point of then given it follows that there exists with hence Since this holds for all Thus Similarly if and there exists with so shows that and so if so and and are separated.

Remark It is not in general the case that and similarly for Make sure you understand why!

d) If is connected and contains at least two points, and then For any there must exist at least one point in with Indeed if not, then is empty and with and as above. Since these sets are separated and shows that neither is empty this would contradict the connectedness of

Chapter 2: Problem 20

Suppose is connected and consider where and Then and so and are separated. Thus one must be empty by the connectedness of changing names if necessary we can assume that it is so must consist only of limit points of and necessarily but then so and is indeed connected.

On the other hand the interior of a connected set need not be connected. Take the region This is the union of the closed first and third quadrants. From the problem below, the two quadrants are themselves connected. Let us show that the union of two connected sets with non-empty intersection is connected. Thus suppose are connected sets and Then if with and separted it follows that with Furthermore, and are separted for (and the same since and From the connectedness of we must have one of or and one of or Of course it we have both empty then and similarly for the 's. So, if necessary swithching and the only danger is that and but then and but then contradicting the assumption. Thus is connected and in particular ths applies to our union of two quadrants.

The interior is the union of the two open quadrants and these are separated, since the closure of each is the corresponding closed quadrant so the union is not connected.

Chapter 2: Problem 21

a) Certainly since and If then it is for is infinite which means that

 (17)

is infinite, so a limit point of hence (since and are separated) so which shows that The same argument shows that b) Put Since this set is non-empty and clearly bounded above, so exists. Moroever, since would imply that Now set Certainly and If then and since otherwise they are not separated so take If then and In any case we have found which implies c) If is convex, then by definition for for From the argument above, if where and are separated then we have a contradiction to the fact that both are non-empty, since then taking and we have found but by convexity. Thus any convex subset of is connected.

Chapter 2: Problem 24

Remark: Note that separable and separated are rather unrelated concepts.

We assume that is a metric space in which every infinite subset has a limit point (sequentially compact is what I called these in lecture). We are to show that is separable. For each choose successively points for such that for Either at some point no further choice is possible, or else we can choose this way an infinite set with for all distinct. By assumption on this set must have a limit point, call it Since must be infinte, it must contain at least two distinct points which have

which is a contradiction. Thus any such procedure must lead to a finite set; call one such choice Then consider This is a countable subset of and by construction for any and any there exists such that (since otherwise we could increase Thus is in the closure of i.e. or is dense. Hence is separable.

Chapter 2: Problem 26 Suppose is sequentially compact in the sense that every infinte subset of has a limit point. We already know from Exercises 23 and 24 that has a countable basis of open sets. That is, there is a countable collection of open sets such that any open set is a union of elements of That is, given open define then Suppose

 (18)

is an arbitrary cover of by open sets. Since each is a union of the sets from if we define

 for some (19)

we must have a subset of hence countable (possibly finite of course). For each we can choose an such that Let be a set chosen in this way, there is a surjective map from to so is countable and by definition of

 (20)

Thus we have found a countable subcover of the original cover.

Now, replace by a labelling by integers, so we can write

 (21)

Of course if is finite we are finished already. Otherwise, suppose that for each

 (22)

The form a decreasing sequence of closed subsets of If they are all nonempty then we can choose a point from each, forming a set Either is finite or else infinite. In the first case there is one point in which is in for arbitrarily large hence in all the since they are decreasing. In the second case must have a limit point, Since, for each all but a finite subset of is contained in must be a limit point of each hence in each (since they are closed). In either case this gives a point in because of (21) this is a contradiction. Thus the must be empty from some point onwards, giving us a finite subcover.

Remark: In lecture I did not go through the step of extracting the countable subcover, just used the cover given by directly.

Chapter 2: Problem 29

Suppose is open. Since is separable, it contains a countable dense subset, for instance Take a surjection and write for the image of Since is open if then for some must contain at least one of the Consider successively each of the Again, there is an interval Now, if consider

 (23)

Similarly if consider

 (24)

In all four cases we obtain an open interval possibly infinite in one direction or the other. By definition this interval is maximal in and containing Another way of thinking about this interval is as the union of all intervals in containing and noting that the union of any collection of open intervals with a fixed point in common is an open interval. Now, drop the 's from which are contained in one of the previous intervals and we have a countable (possibly finite) collection of disjoint intervals with union (since each point of is in an interval containing one of the

Richard B. Melrose 2004-05-18