Chapter 2: Problem 19
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
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
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
Now, replace by a labelling by integers, so we can write
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
Richard B. Melrose 2004-05-18