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.
Triangle inequality (In R2)
Probably the most important inequality in this course.
∣x+y∣≤∣x∣+∣y∣
Proof
First, we prove the claim: ∣x∣≤a⟺−a≤x≤a
Here I only prove ‘‘⇒"direction.The other direction can be easily obtained by "opening" the absolute value.⇒⇒Case1. x≥0x=∣x∣≤a,x≥0⇒a≥0⇒−a≤0≤x−a≤x≤aCase2. x<0−x=∣x∣≤a⇒x≥−a⇒x<0≤a−a≤x≤aFor the triangle inequality:
∣x∣∣y∣−∣x∣−∣y∣⇒−(∣x∣+∣y∣)⇒∣x+y∣≤∣x∣⇒−∣x∣≤x≤∣x∣≤∣y∣⇒−∣y∣≤y≤∣y∣≤x+y≤∣x∣+∣y∣≤x+y≤∣x∣+∣y∣≤∣x∣+∣y∣ by the claim above □Notes
The very obvious ∣x∣≤∣x∣ may be hard to come up with. When you have an equality, think about inequalities it may give you. The generalization for Rn can be proved by induction.
Q is dense in R
That is, for any x,y∈R,x<y,∃q∈Q such that x<q<y . Equivalently, there are infinite rational numbers between any two real numbers.
Proof
Let n∈N,0<n1<y−x⟺n>y−x1 by the Archimedean property.
Consider the set S={k∈N:nk<x}. This set is finite, so its complement is infinite, non-empty especially. By the well-ordering principle, the set R∖S={k∈N:nk>x} has a least element.
Let m be the smallest natural number with nm>x, especially nm−1<x. m,n∈N,nm∈Q.
Claim: x<nm≤y is the number we want.
⇒nm=nm−1+n1<x+y−x=ynm>x by our choice of m x<nm<y □Notes
The choice of n1<y−x in the first step seems mysterious, but everything follows that is easy to understand. The reason why we made this choice is: n1 is the step size, if it’s greater than y−x, assume z<x,z+n1 may be greater than y. That is, we may go past y directly, which is bad for finding a number between x,y.
Of course, the set S={k∈N:nk<x} may be empty ( n1>x ), but that doesn’t matter. R∖S is “more” infinite. The well-ordering principle holds as long as R∖S 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
Every continuous function on a compact domain is uniformly continuous.
Proof
Let f:A→R be continuous. Let ϵ>0, then ∀ x∈ A, ∃δx>0:∣x−u∣<δx⟹ ∣f( x) −f( u)∣ < 21ϵ (1)
Consider the neighborhoods V21δx(x), ∀x∈A. Then, C :={V21δx(x):x∈A} is an open cover of A since an arbitrary union of open sets are still open. A is compact, so C has a finite subcover, U:={V21δxk(x1), ⋯,V21δxk(xk)} where x1,⋯,xk∈A
Let x, u ∈A be arbitrary. We want to show ∣x−u∣<δ⟹∣f(x)−f(u)∣<ϵ. For now, we only know f behaves nicely locally(within each neighborhood around xi, 1⩽i⩽k ), but uniformly continuity is a global property.
Intuition: If x,u are in the same neighborhood around xi, we can utilize the nice, local behavior. The problem is that δxi depends on xi, so the required distance varies in different locations on the domain. Can we find a δ that makes x and u “close enough” to be considered locally, no matter where they are?
Yes, we can. Why? Because U is finite.
Let δ:=min{21δx1,⋯,21δxk}. U covers A, so x∈V21δxi(xi)for some xi∈A.
∣x−xi∣∣u−xi∣triangle ineq∣u−xi∣<δxi∣f(x)−f(u)∣<21δxi<δxi⟹∣f(x)−f(xi)∣<21ϵ by (1)=∣u−x+x−xi∣≤∣u−x∣+∣x−xi∣<δ+21δxi≤21δxi+21δxi=δxi⟹∣f(u)−f(xi)∣<21ϵ=∣f(x)−f(xi)+f(xi)−f(u)∣≤∣f(x)−f(xi)∣+∣f(xi)−f(u)∣<21ϵ+21ϵ=ϵ □Notes
If U is infinite, we are unable to take the minimum of all neighborhoods. To recap, our goal is to find a δ that is “close enough” for arbitrary x, u ∈A such that ∣f(x)−f(u)∣<ϵ. The idea is to use xi as a bridge between x and u since we only know how f behaves near each xi. Choosing 21δxineighborhoods ensures x, u are in the same neighborhood of xi. This enables us to rely on f’s local behavior.
Bolzano-Weierstrass Theorem: Every bounded sequence in R has a convergent subsequence
The proof is only for R, but the theorem also holds in Rn and Cn.
Definition: Let (xn) be a sequence of real numbers. xN is called a peak if ∀n≥N,xN≥xn i.e., xN is the largest term in the sequence starting from it.
Lemma: Every sequence of real numbers has a monotone subsequence
Proof
Let (xn) be a sequence of real numbers. There are two cases regrading the number of peaks:
Case1: (xn) has infinitely many peaks. We can find a subsequence (xnk) consisting of peaks. (xn1) is a peak, so xn1≥xn,∀n≥n1. Thus, xn1≥xn2. Similarly, xn2 is a peak, so xn2≥xn3≥xn4≥⋯, etc.
The sequence is monotone increasing.
Case2: (xn) has finitely many peaks. Then, we choose our subsequence after the last peak. Let xN be the last peak. For xn,n≥N,∃xm,m>n,such that xm>xn since xn is not a peak. Similarly, xm is not a peak, so ∃xk,k>m,such that xk>xm. Following this procedure, we end up with a monotone increasing subsequence.
In both cases, there exists a monotone subsequence in (xn), so the lemma is proved.
Proof of the theorem
(xn) is bounded. By lemma, (xn) has a monotone subsequence (xnk). By the monotone convergence theorem, the sequence converges □.
Notes
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