9:19 PM

Let's look again at the theorem from last Thursday, specifically as applied to subsets. Suppose K_{1} is a compact set. Look at a closed subset of K_{1}. We know that any closed subset of a compact set is itself compact, so we can call the subset K_{2}.

We can generalize this to K_{n} as a compact set and K_{n+1} as a compact subset of K_{n}, for any natural number n, giving us an infinite sequence of compact subsets. If we specify that none of the subsets are empty, then last Thursday's result states that the intersection of all of the subsets, ∩_{n∈ℕ} K_{n}, is not empty.

On the one hand, this seems like it should be obvious. On the other hand, we had an example of a sequence of closed subsets yesterday where this is not true. It's also easy to construct a sequence of open subsets where this is not true. Let E_{n} be the open segment from 0 to 1/n on the real number line. (For example, E_{4} = all real numbers x such that 0<x<1/4.) Then for any natural number n, E_{n} is not empty. Also, for any any natural number n, E_{n+1} is a subset of E_{n}. However, if you pick any real number greater than 0, there always exists a natural number n such that 1/n is less than the real number. Therefore, that real number is not a member of E_{n}, so the intersection of all possible E_{n} must be empty.

After looking at some examples of sequences of non-empty subsets where the intersection of the subsets is empty, it's no longer obvious that the intersection of a particular sequence of subsets must be non-empty. So we've actually proved something useful. If the sets are compact, what we naively expect to happen does in fact happen, and the intersection of the sequence of non-empty compact subsets is itself non-empty.

Today's goal is to show that the same thing is true for closed intervals on the real number line. We have not shown (yet) that closed intervals are in fact compact sets, so we will have to approach the problem from a different direction.

Suppose I_{1} is a closed interval on the real number line. That is, for two real numbers a_{1} and b_{1}, where a_{1}≤b_{1}, I_{1} is the set of real numbers x such that a_{1}≤x≤b_{1}. Next, suppose I_{2} is a closed interval which is a subset of I_{1}. Then I_{2} is the set of real numbers x where a_{2}≤x≤b_{2}. Since I_{2} is a subset of I_{1}, a_{1}≤a_{2}≤b_{2}≤b_{1}.

We can generalize to I_{n} for any natural number n, and say that I_{n+1} is a subset of I_{n}, so I_{n}={x∈ℝ: a_{n}≤x≤b_{n}} and a_{n}≤a_{n+1}≤b_{n+1}≤b_{n} for all natural numbers n.

Now, look at the set of all a_{n}. This set is bounded above, since a_{n}≤b_{1} for all n. Therefore, the set has a least upper bound, or supremum. Call the supremum x. The claim is that x≤b_{n} for all n. Assume there is some number N such that x>b_{N}. Then there exists a number y between them, so x>y>b_{N}. Now, for all n≥N, a_{n}≤b_{n}, and b_{n}≤b_{N}. This means that y>a_{n} for all n, so y is also an upper bound for the set of all a_{n}. But x is the least upper bound, so y cannot exist. Therefore, x≤b_{n} for all n.

This means that for all n, a_{n}≤x≤b_{n}. Therefore, x is a member of every I_{n}, so x is a member of the intersection of all the I_{n}. So we have shown what we set out to prove. Given an infinite sequence of closed intervals on the real number line, such that each interval is a subset of the previous interval, the intersection of all of the intervals contains at least one point.

I just want to compare the two proofs. Looking at the compact sets, the proof is dependent on lots of architecture of metric spaces. We have to use the definition of compact sets and the relationship between open sets and closed sets, and do a bunch of juggling back and forth between the different properties.

Looking at closed intervals on the real number line, the entire argument comes down to the facts that the real numbers are ordered, and there is always a real number between any two real numbers. This strikes me as a much simpler argument, in that it requires much less of a framework. You don't have to worry about concepts like distance functions or neighborhoods, never mind higher level abstractions like open, closed, and compact sets.

However, we end up with the same basic conclusion in both cases. Given a particular collection of sets, where the sets are subsets of each other, the intersection of the sets is not empty. I'm not sure if there's a profound insight here, or if all we've done is place another stepping stone toward tomorrow night's, and next week's, goals.

10:46 PM

I want to clean up last week's post about infinite intersections of compact sets. But first I want to complain about the presentation of the material on compact sets in Principles of Mathematical Analysis. This book, or at least this chapter of the book, is significantly short on two things: motivation, and examples.

I'm sure Rudin knows why he's presenting all of this material here. I assume that in a few chapters, he'll be discussing continuity and say something like, "this set is compact, so it has this useful property, which allows us to conclude this result," and I'll look at it and think that the result looks like it might be kind of interesting and so I'm glad I spent so much time on compact sets. Except then I probably won't understand why the result is actually important, because he won't provide any motivation there either. And the way it's written will probably be closer to "This result follows from Theorem 2.46" and I will flip back to the theorem and have to figure out why it's relevant to the current situation. Rudin is a man of few words and doesn't feel a need to make things easy for the reader.

It would be easier to figure out what's actually important if I had a sense of where things were going. To a certain extent, it can be easier to follow a chain of mathematical reasoning if you start at the conclusion and work backward from there. This can end up fragmenting the thought process, so it may be necessary to develop all the pieces in reverse order, and then reassemble them in forward order so the whole chain of thinking becomes clear. Right now I don't know what the end goal is, so I'm just hanging on and hoping I pick up the right details so it makes sense when I get there.

In my blogging, I have a short term goal I'm working toward. I'm not saying what it is, so I'm probably as guilty as Rudin about not providing motivations. I know I can get there, but I don't yet have the full picture in my head, so I'm hoping that when I do get there, I will have already included all the necessary details. Once I reach this goal, I plan to move on to a different topic. I currently expect to get there some time next week, but last week, I thought I would get there this week, so it's turning into a moving target.

The second thing Rudin doesn't really bother to provide is examples. He provides the definition of compact sets, and it involves infinite collections of sets. Any time you have to talk about infinite anything in order to define something, actually understanding the definition is hard work. So it would be great to have some concrete examples to make sense of the definition. Instead Rudin launches straight into theorems about compact sets, and we're quickly looking at theorems involving infinite collections of compact sets, and this is way too much infinity to think about.

Which brings me back to Thursday's post. I set out to provide an example of an infinite collection of closed but not compact sets. Unfortunately, the sets in my example were not closed. My sets were based on two rational numbers p and q, taking all rational numbers x such that p ≤ x ≤ q, on the real number line. A moment's thought leads to the conclusion that these may be closed sets on the rational number line, but they are not closed on the real number line. The definition of a closed set is that every limit point of the set is a member of the set. Any irrational number between p and q is a limit point of this set. You could say that this is because the rational numbers are dense in the real numbers. But it also can be shown pretty directly from the definitions of limit points and closed sets. Since the irrational numbers are excluded from the sets as I defined them, they are not closed.

So let's start over, with a new collection of sets, which are actually closed. Pick a natural number n. On the real number line, take the set of all real numbers x such that x ≥ n. (Just a reminder that as I've been using it, the real number line is the set of real numbers with distance function d(p,q) = |p−q|. The real number line is a metric space.) The set of real numbers greater than or equal to n is a closed set. No point less than n can be a limit point of the set, and the set includes all points greater than or equal to n, so it must be closed.

That's one set. Now, take the collection of all sets with every starting value of n. We can number them, and say that set A_{1}={x∈ℝ: x≥1}, A_{2}={x∈ℝ: x≥2}, and in general A_{n}={x∈ℝ: x≥n} for any natural number n. Since the set of natural numbers is infinite, the collection of sets is infinite.

Now, take some finite collection of these sets, for example, A_{1}, A_{5}, and A_{17}. The real number 2 is in the first set, but not the other two. However, the real number 18 is in all three sets. 18 is a point in the intersection of the three sets, so the intersection is not empty. In general, in any finite collection of sets, there will be a set with a maximum starting point, and all real numbers greater than or equal to that starting point will be in the intersection of the sets. Any intersection of a finite collection of these sets is non-empty.

But let's look at the infinite collection of all the sets. For any real number x, there exists a natural number n greater than x. A_{n} is one of the sets in the infinite collection, and x is not a member of A_{n}, so x is not a member of the intersection of the infinite collection. This is true for any real number x, so the intersection of the infinite collection of sets is empty.

However, A_{n} is closed, but it is not compact. To show that, let's construct an infinite open cover of A_{n}, where if we remove any set in the open cover, it will no longer cover A_{n}. Therefore no finite subcover of the infinite open cover exists, and A_{n} is not compact. For each natural number greater than or equal n, construct an neighborhood of radius 3/4 centered on that number. (The radius has to be greater than 1/2 so all the sets overlap, but it has to be less than or equal to 1 so they don't overlap too much.) The neighborhoods are open sets, each covering part of A_{n}. It should be clear that every real number in A_{n} is covered by the open set centered on the nearest natural number, so the collection of all the neighborhoods is an open cover of A_{n}. Since the set of natural numbers greater than or equal to n is an infinite set, the collection of open sets is infinite. If any set in the cover is removed, the number the set was centered on will not be covered by any other set, so no finite subcover exists. Therefore, A_{n} is not compact.

My example from Thursday also failed because we haven't shown yet that closed intervals on the real number line are compact sets. We will show that soon, probably at the beginning of next week, but until then I should use a different example, of sets that we have already shown are compact. So far, we have one example, from the post where I defined compact sets. This is the set of all rational numbers of the form 1/n, where n is a natural number, and the number 0.

We can turn this into a collection of compact sets. Call K_{N} the set of all rational numbers of the form 1/n, where n is a natural number greater than or equal to N, and the number 0. So K_{1} is the original set, and as N increases, we chop off points from the right side of the set. Now, for any finite collection of K_{N}, there's a maximum N, and all points of the form 1/n, where n is greater than or equal to that maximum N, are in the intersection of all the sets.

If we look at the collection of all of the sets K_{N}, for any number 1/n, there is an N greater than n. Since K_{N} is a member of the collection, no point of the form 1/n is in the intersection of the infinite collection. (Question, in case anyone is still reading: is this sufficient, or should I include an example with actual numbers substituted in for the n's and N's? In general, I'm unsure of the level of specificity I should use in my examples to make them comprehensible.) However, every set in the collection also includes the number 0, so the intersection of the infinite collection of sets consists of a single point, 0.

Revisiting my conclusion from last time, every one of the closed sets A_{n} is an infinite set. And the intersection of a finite collection of these sets is an infinite set. In a finite collection, there are points which are in some of the sets but not in the intersection of the sets, so in that sense the intersection is smaller. This can be misleading, since the sets are all infinite and the cardinality of the sets is the same, so looking at the cardinality, all of the sets are the same size. In any event, the intersection of all of the infinite sets is empty.

In comparison, the compact sets K_{n} are also infinite sets. The intersection of a finite collection of these sets is also infinite. However, the intersection of the infinite collection of sets is not empty. In this case, it is the single point 0. We could also construct collections of compact sets such that the intersection of the infinite collection is any finite size, or any infinite set. The point of the theorem from last time is that unlike with the closed sets, with compact sets the intersection of the infinite collection will never be empty.

One of my goals with my examples from Thursday was that the sets not be subsets of each other. If the compact sets are labeled by K_{α}, the values of α don't even have to be ordered, but if they are, K_{n+1} does not have to be a subset of K_{n}. I abandoned that goal for my example this time, and the compact sets I've used are in fact all subsets of other sets in the collection. This is a useful special case of the theorem, but it is not a requirement of the theorem. Now that I have a working example, I can only hope that it is not overly specific.