9:19 PM

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 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, E₄ = 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 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 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≤b₁ 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.

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