Total Marks possible:
Average Mark: 43
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.
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.
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.
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.
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
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.
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