wherein is detailed Matt's experiences as he tries to figure out what to do with his life. Right now, that means lots of thinking about math.

Wednesday, April 11, 2012

Let's look again at the theorem from last Thursday, specifically as applied to subsets. Suppose K1 is a compact set. Look at a closed subset of K1. We know that any closed subset of a compact set is itself compact, so we can call the subset K2.

We can generalize this to Kn as a compact set and Kn+1 as a compact subset of Kn, 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 Kn, 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 En be the open segment from 0 to 1/n on the real number line. (For example, E4 = all real numbers x such that 0<x<1/4.) Then for any natural number n, En is not empty. Also, for any any natural number n, En+1 is a subset of En. 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 En, so the intersection of all possible En 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 I1 is a closed interval on the real number line. That is, for two real numbers a1 and b1, where a1≤b1, I1 is the set of real numbers x such that a1≤x≤b1. Next, suppose I2 is a closed interval which is a subset of I1. Then I2 is the set of real numbers x where a2≤x≤b2. Since I2 is a subset of I1, a1≤a2≤b2≤b1.

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

Now, look at the set of all an. This set is bounded above, since an≤b1 for all n. Therefore, the set has a least upper bound, or supremum. Call the supremum x. The claim is that x≤bn for all n. Assume there is some number N such that x>bN. Then there exists a number y between them, so x>y>bN. Now, for all n≥N, an≤bn, and bn≤bN. This means that y>an for all n, so y is also an upper bound for the set of all an. But x is the least upper bound, so y cannot exist. Therefore, x≤bn for all n.

This means that for all n, an≤x≤bn. Therefore, x is a member of every In, so x is a member of the intersection of all the In. 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.

FAQ

What does "rolls a hoover" mean, anyway?

"Roll a hoover" was coined by Christopher Locke, aka RageBoy (not worksafe). He enumerated some Hooverian Principles, but that might not be too helpful. My interpretation is that rolling a hoover means doing something that you know is stupid without any clear sense of what the outcome will be, just to see what will happen. In my case, I quit my job in an uncertain economy to try to start a business. I'm still not sure how that will work out.