I’ve encountered a lot of proofs from MATH254, an introductory real analysis course. From my experience, most of the knowledge vanished the moment I walked out of the final exam room. But some proofs are just so elegant that forgetting them pains me.
This post is a collection of the neatest ones from my perspective. Besides reconstructing proofs, I also try to explain them in my own words to help you and my future self follow the line of thought.
Probably the most important inequality in this course.
First, we prove the claim:
For the triangle inequality:
The very obvious may be hard to come up with. When you have an equality, think about inequalities it may give you. The generalization for can be proved by induction.
That is, for any such that . Equivalently, there are infinite rational numbers between any two real numbers.
Let by the Archimedean property.
Consider the set . This set is finite, so its complement is infinite, non-empty especially. By the well-ordering principle, the set has a least element.
Let be the smallest natural number with , especially . .
Claim: is the number we want.
The choice of in the first step seems mysterious, but everything follows that is easy to understand. The reason why we made this choice is: is the step size, if it’s greater than , assume may be greater than . That is, we may go past directly, which is bad for finding a number between .
Of course, the set may be empty ( ), but that doesn’t matter. is “more” infinite. The well-ordering principle holds as long as is non-empty. Using infinity to show non-emptiness may seem like an overkill, but interestingly, this kind of “overkill” happens all the time. It’s easier to get information from a stronger statement.
We require f to be continuous on the whole domain, but actually we only used continuity at a single point in this proof. — My analysis professor
Let be continuous. Let then (1)
Consider the neighborhoods . Then, is an open cover of A since an arbitrary union of open sets are still open. A is compact, so has a finite subcover, where
Let be arbitrary. We want to show . For now, we only know behaves nicely locally(within each neighborhood around ), but uniformly continuity is a global property.
Intuition: If are in the same neighborhood around , we can utilize the nice, local behavior. The problem is that depends on , so the required distance varies in different locations on the domain. Can we find a that makes and “close enough” to be considered locally, no matter where they are?
Yes, we can. Why? Because is finite.
Let . covers A, so for some .
If is infinite, we are unable to take the minimum of all neighborhoods. To recap, our goal is to find a that is “close enough” for arbitrary such that . The idea is to use as a bridge between and since we only know how behaves near each . Choosing neighborhoods ensures are in the same neighborhood of . This enables us to rely on ’s local behavior.
Every bounded sequence in has a convergent subsequence.
The proof is only for , but the theorem also holds in and .
Definition: Let be a sequence of real numbers. is called a peak if i.e., is the largest term in the sequence starting from it.
Let be a sequence of real numbers. There are two cases regrading the number of peaks:
Case1: has infinitely many peaks. We can find a subsequence consisting of peaks. is a peak, so . Thus, . Similarly, is a peak, so , etc.
The sequence is monotone increasing.
Case2: has finitely many peaks. Then, we choose our subsequence after the last peak. Let be the last peak. For since is not a peak. Similarly, is not a peak, so . Following this procedure, we end up with a monotone increasing subsequence.
In both cases, there exists a monotone subsequence in , so the lemma is proved.
is bounded. By lemma, has a monotone subsequence . By the monotone convergence theorem, the sequence converges .
The proof is elegant and more importantly, easy to follow. The definition of peak is the key observation which makes the proof intuitive.
The power of a perfect definition. — The Teaching assistant