# Solutions to first Test

Total Marks possible:

Average Mark: 43

Median: 40

You are permitted to bring the book `Rudin: Principles of Mathematical Analysis' with you - just the book, nothing else is permitted (and no notes in your book!) You may use theorems, lemmas and propositions from the book. Note that where is mentioned below the standard metric is assumed.

1. Suppose that is a sequence in a metric space, and Assuming that every subsequence of itself has a subsequence which converges to show that

Solution:- To say that converges to is to say that for every the set is finite. Thus if were not to converge to then for some this set would be infinite. Then we can take the subsequence where is the unique increasing sequence with range Thus sequence cannot have any subsequence converge to since for any susbsequence (of the subsequence) all points lie outside This proves the result by contradiction.

2. Let be a sequence of real numbers with Show that there is a subsequence such that

Solution:- By definition of convergence, give for every there exists such that implies Thus we can choose a subsequence with for all Then so the sequence of partial sums is increasing and bounded above, hence convergent.

3. Give examples of:
1. A countable subset of which is infinite and closed.
2. A subset of the real interval which contains but is not compact.
3. A metric space in which all subsets are compact.
4. A cover of as a subset of which has no finite subcover.

Solution:- Note that I did not ask you to prove this, but they do need to be right. Not just discrete in the third case, etc.

1. The subset is closed since the only limit point is It is certainly infinite and countable.
2. For example contains but is not compact since it is not closed.
3. Any finite metric space has the property that all subsets are finite, hence compact.
4. Taking the collection of single-point subsets is a cover of but has no finite subcover. (I did not say open).

4. Let be a compact set in a metric space and suppose Show that there exists a point such that

Solution:-

Method A

By definition of the infimum, there exists a sequence in such that Since us compact, this has a convergent subsequence. Replacing the original sequence by this subsequence we may assume that in Now, by the triangle inquality

by the definition of convergence. Thus must be the limit of the sequence in so as desired.
Method B
Let If there is no point with then the open sets cover By compactness there is a finite subcover, so for some which contradicts the definition of
Method C
(really uses later stuff) Since is continuous (prove using triangle inequality) and is compact, is compact in so contains its infimum. Thus there exists with

1. Let be a (non-empty) metric space with metric and let be an external point. Show that there is a unique metric one satisfying

2. Show that with this metric is not connected.

Solution:- Unfortunately I got carried away here and this is not true! I should have said that is a metric space with for all then it works fine. I hope I did not confuse anyone too much by this. I gave everyone full marks for the whole question.

5. Let be a metric space and Let be the union of all those open sets in which are subsets of Show that the complement of is the closure of the complement of

Solution:- A set contained in is exactly one with complement containing the complement of Thus, from the definition, the complement of is the intersection of all closed sets which contain the complement of This we know to be its closure.

Or:- - since a point in is either an interior point (lies in an open subset of or else is a limit point of the complement). Thus which is the closure of the complement.

Richard B. Melrose 2004-05-18