So here is one proof for the question from the previous post.  In case you forgot I shall now restate it:

Assume that in a metric space you are given 2 closed disjoint sets and .  Prove that if one of them is compact then with and .

proof: Without loss of generality (WLOG) assume that is compact.  Then define as .  Otherwise, define it to be the infimum of the distance from the point in to a point in the set .  Now is continuous and thus attains its infimum.  But this must be positive, because if it were not we would have contradicted the fact that the sets were disjoint and closed.

This proof works, but I would also like to find one that relies on a set being compact if any open cover has a finite subcover.  Sure it might be more work than necessary, but I feel that it should be good exercise.

I was on a math help forum and was trying to help someone with the following problem.

Given two disjoint closed sets and in a metric space , prove that there exists disjoint open sets and such that and .

In my proof I made the following mistake: Let for , . Then .

I should have stopped at this time, since I remember talking about something very similar to this in my analysis class.  It turns out that to gaurantee you need that at least one of to be compact.

Here is an interesting problem that I saw a few days ago.  I don’t really know why I found this more interesting that other problems/solutions I’ve seen but it just is.

Here is the problem statement (generalized).  Consider a function , and ,  such that for any sequence .  Prove that the set of points where is not zero is countable.

Proof:

Suppose for the sake of contradiction that the set is not countable, thus uncountable.  Define .  It should be obvious that . So one of the must be uncountable.  Thus if we choose points from this set (one of the uncountable sets) we can make as large as we want. Hence we must have that the set of points wehre is at most countable.

Not that in fact we can ‘weaken’ the assumption on so that it reads .

Well here is a question that I used to wonder about for a while.

If is integrable over , then

Well here is a proof of a simplified version.  I only assume that and

proof: By “the world’s most obvious integral inequality” So obviously

For the reverse direction let then on some set Therefore To get the result take then .

The only part that bothers me is making sure that is measurable, but that shoudn’t be too difficult to fix.

Well I got my Putnam results back today. I managed to score a 20 (which is double my score from last year). I was hoping to make the top 15% but just missed and am instead in the top 17%. In addition to this I missed being in the top 500 by 119 places (which is probably the equivalent of 1 or 2 points).

Hopefully next year I’ll get top 500.

Well I haven’t posted anything here in a while. The main reason is that it is currently winter break and I am using the time to do other things. For instance I am finally sitting down to solve some SPOJ problems I have marked as interesting or “useful”.

Well it has been a while since I have metioned my math adventures.  A few days ago I took the Putnam exam. Personally I found the morning seesion to be kind of “easy” ( for the Putnam that is). I was able to answer 2 questions completely and half of a third. The afternoon session was more of what I expected. I knew the answer to 2 of the questions but couldn’t think of a proof for one of them, and then the other I was making a computational mistake (how depressing).

I’m not going to post the problems since it shouldn’e be too difficult to find them with a simple Google search or looking on the AoPS fourms.

After talking to my school mates I felt that I could have answered 4 of the morning session questions, if I didn’t have a thing against using the integral test. My favorite problem on the whole test was as follows: What is the maximum number of rational points that can be on a circle with a center who’s is irrational.

Oh well I am actually pretty happy with my 20, an improvement from last year’s 10, who knows maybe I’ll get a 21 but that’s unlikely.

Well we all know that the only solutions to are and where . So let us move onto something a little more “difficult”. Find all integer triplets satisfying the following equation.

We begin by doing the obvious thing and expand the right hand side and then cancel terms, thus leaving us with . Then divide by 3 to get , which after some simple regrouping factors into . From here it is obvious that the only solutions are those where one of is euqal to the negation of one of the others. See pretty simple, the only real work is being able to expand the first expression. The factoring might be a little tricky but just from looking at the equation you should have a general idea of what the solution should be and that should help you get pretty far with the factoring.

(a) Since is continuous on a compact set ( it attains both its maximum and minimum values, so for all . Now using these poor estimates we have that . So if we consider the function we have that is continuous on the same interval as . Thus we just apply the intermediate value theorem to and the desired result follows.

(b) WLOG assume that . Let be given and define new function by . This is just the equation for the slope of the line connecting the points and and is continuous because is continuous. Now consider the set of points for . Now look at the values takes at these points. If for any then we have found a solution, so assume that for any chose of . So WLOG assume that , since we know that there must be some such that , call this . Then we there is a such that .

(c) For this one I will prove something a little stronger. All that we need to assume about is that it is integrable, and has what I shall call the extreme value property on our interval. Basically this property says that on the interval there exists a point such that and a point such that , where ranges over all the values in our interval. So clearly an increasing function satisfies these conditions. Now define a function by

Now is continuous and , thus by the intermediate value theorem we are done.

Some after thoughts: Problem (a) was pretty straight forward and didn’t take too much insight to actually solve. However, problem (b) was a little more challenging until I visualized what I was being asked to prove. This led me to the idea to look at the slope of the connecting line segment. Finally problem (c), I at first wondered why you were given that was increasing and not continuous, or even just integrable. As can be seen in my proof continuity imposes more conditions than are necessary, while integrability does not provide you with enough. This led me to think about what makes increasing functions special (well one of the things that makes them special).

Well I am in a big Analysis mood. I guess that it has something to do with the fact that at this point in my life Analysis is the topic in math that spikes my interest the most. This can be accredited to my Calculus BC teacher from high school (Mrs. Johnson) and my undergraduate Advanced Calculus professor (Dr. Shen). They both did a wonderful job of presenting material and showing applications and fascinating consequences.

Anyway onto the post, I was reading through the intermediate real analysis section of Problem Solving Through Problems. Here are some of the problems I found interesting.

(a) Suppose that is continuous and is integrable and such that for all . Prove that there is a number in such that

(b) Let be continuous and suppose that . Prove that for each positive integer there is an in such that .

(c) Assume the same conditions on as in part (a) except that instead of being continuous is now assumed to be increasing. Prove that there is a such that

My thoughts: The solution to (a) is a pretty straight forward application of the intermediate value theorem. For problem (b) I am considering looking at the slope of the line connecting and . Finally for (c) I am going to try and generalize it in an appropriate way. My solutions should be posted in the near future.