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Sixian Li

My favorite proofs from Analysis1

Math

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\mathbb{R}^2)

Probably the most important inequality in this course.

x+yx+y|x+y|\leq|x|+|y|

Proof

First, we prove the claim: xa    axa|x|\leq a \iff -a\leq x\leq a

Here I only prove "direction.The other direction can be easily obtained by "opening" the absolute value.Case1. x0x=xa,x0a0a0xaxaCase2. x<0x=xaxax<0aaxa\text{Here I only prove } ``\Rightarrow" \text{direction.}\\ \text{The other direction can be easily obtained by "opening" the absolute value.}\\ \begin{aligned} &\text{Case1. } x\geq 0 \\ &x = |x|\leq a, x\geq0 \Rightarrow a\geq0\Rightarrow -a \leq 0\leq x\\ \Rightarrow &-a \leq x \leq a\\ &\text{Case2. } x\lt 0 \\ & -x =|x|\leq a \Rightarrow x\geq-a\Rightarrow x<0\leq a\\ \Rightarrow &-a \leq x \leq a\\ \end{aligned}\\

For the triangle inequality:

xxxxxyyyyyxyx+yx+y(x+y)x+yx+yx+yx+y by the claim above \begin{aligned} |x|&\leq |x| \Rightarrow -|x| \leq x\leq|x|\\ |y|&\leq |y| \Rightarrow -|y| \leq y\leq|y|\\ -|x|-|y|&\leq x+y \leq|x|+|y|\\ \Rightarrow -(|x|+|y|)&\leq x+y \leq|x|+|y|\\ \Rightarrow |x+y| &\leq |x|+|y| \text{ by the claim above} \ \square \end{aligned}

Notes

The very obvious xx|x|\leq|x| may be hard to come up with. When you have an equality, think about inequalities it may give you. The generalization for Rn\mathbb{R^n} can be proved by induction.

Q\mathbb{Q} is dense in R\mathbb{R}

That is, for any x,yR,x<y,qQx, y \in \mathbb{R}, x<y, \exists q \in \mathbb{Q} such that x<q<yx < q < y . Equivalently, there are infinite rational numbers between any two real numbers.

Proof

Let nN,0<1n<yx    n>1yxn\in\mathbb{N},0<\frac{1}{n} < y-x \iff n>\frac{1}{y-x} by the Archimedean property.

Consider the set S={kN:kn<x}S = \{ k \in \mathbb{N}: \frac{k}{n}\lt x \}. This set is finite, so its complement is infinite, non-empty especially. By the well-ordering principle, the set RS={kN:kn>x}\mathbb{R}\setminus S = \{k\in\mathbb{N}:\frac{k}{n}>x\} has a least element.

Let mm be the smallest natural number with mn>x\frac{m}{n} > x, especially m1n<x\frac{m-1}{n}< x. m,nN,mnQm, n \in \mathbb{N}, \frac{m}{n} \in \mathbb{Q}.

Claim: x<mnyx<\frac{m}{n}\leq y is the number we want.

mn=m1n+1n<x+yx=ymn>x by our choice of m x<mn<y \begin{aligned} &\frac{m}{n} =\frac{m-1}{n} + \frac{1}{n} < x+y-x =y\\ &\frac{m}{n} >x \text{ by our choice of m}\\ \Rightarrow &\ x<\frac{m}{n}<y \ \square \end{aligned}

Notes

The choice of 1n<yx\frac{1}{n}<y-x in the first step seems mysterious, but everything follows that is easy to understand. The reason why we made this choice is: 1n\frac{1}{n} is the step size, if it’s greater than yxy-x, assume z<x,z+1nz<x, z+\frac{1}{n} may be greater than yy. That is, we may go past yy directly, which is bad for finding a number between x,yx, y.

Of course, the set S={kN:kn<x}S =\{k\in\mathbb{N}:\frac{k}{n} \lt x\} may be empty ( 1n>x\frac{1}{n}\gt x ), but that doesn’t matter. RS\mathbb{R}\setminus S is “more” infinite. The well-ordering principle holds as long as RS\mathbb{R}\setminus 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:ARf: A→ \mathbb{R} be continuous. Let ϵ>0,\epsilon >0, then  x A, δx>0:xu<δx f( x) f( u) < 12ϵ\forall \ x\in \ A,\ \exists \delta _{x} >0:|x-u|< \delta _{x} \Longrightarrow \ |f( \ x) \ -f( \ u) |\ < \ \frac{1}{2} \epsilon (1)

Consider the neighborhoods V12δx(x), xAV_{\frac{1}{2} \delta _{x} }( x) ,\ \forall x\in A. Then, C :={V12δx(x):xA}C\ :=\{V_{\frac{1}{2} \delta _{x} }( x) :x\in A\} is an open cover of A since an arbitrary union of open sets are still open. A is compact, so CC has a finite subcover, U:={V12δxk(x1), ,V12δxk(xk)}U:=\{V_{\frac{1}{2} \delta _{x_{k} }}( x_{1}) ,\ \cdots ,V_{\frac{1}{2} \delta _{x_{k} }}( x_{k})\} where x1,,xkAx_{1} ,\cdots ,x_{k} \in A

Let x, u Ax,\ u\ \in A be arbitrary. We want to show xu<δf(x)f(u)<ϵ|x-u|< \delta \Longrightarrow |f( x) -f( u) |< \epsilon. For now, we only know ff behaves nicely locally(within each neighborhood around xi, 1ikx_{i} ,\ 1\leqslant i\leqslant k ), but uniformly continuity is a global property.

Intuition: If x,ux, u are in the same neighborhood around xix_{i}, we can utilize the nice, local behavior. The problem is that δxi\delta _{x_{i} } depends on xix_{i}, so the required distance varies in different locations on the domain. Can we find a δ\delta that makes xx and uu “close enough” to be considered locally, no matter where they are?

Yes, we can. Why? Because UU is finite.

Let δ:=min{12δx1,,12δxk}\delta :=\min\left\{\frac{1}{2} \delta_{{x}_{1}},\cdots ,\frac{1}{2} \delta_{{x}_{k}}\right\}. UU covers A, so xV12δxi(xi)x \in V_{\frac{1}{2} \delta_{{x}_{i}}}( x_{i})for some xiAx_{i} \in A.

xxi<12δxi<δxif(x)f(xi)<12ϵ by (1)uxi=ux+xxitriangle inequx+xxi<δ+12δxi12δxi+12δxi=δxiuxi<δxif(u)f(xi)<12ϵf(x)f(u)=f(x)f(xi)+f(xi)f(u)f(x)f(xi)+f(xi)f(u)<12ϵ+12ϵ=ϵ \begin{gathered} \begin{aligned} |x-x_{i} | & < \frac{1}{2} \delta _{ {x}_{i} } < \delta _{ {x}_{i} } \Longrightarrow |f( x) -f( x_{i}) |< \frac{1}{2} \epsilon \ \text{by (1)}\\ |u-x_{i} | & =|u-x+x-x_{i} |\\ \text{triangle ineq} & \leq |u-x|+|x-x_{i} |< \delta +\frac{1}{2} \delta _{ {x}_{i} } \leq \frac{1}{2} \delta _{ {x}_{i} } +\frac{1}{2} \delta _{ {x}_{i} } =\delta _{ {x}_{i} }\\ |u-x_{i} |< \delta _{ {x}_{i} }& \Longrightarrow |f( u) -f( x_{i}) |< \frac{1}{2} \epsilon \\ |f( x) -f( u) | & =|f( x) -f( x_{i}) +f( x_{i}) -f( u) |\\ & \leq |f( x) -f( x_{i}) |+|f( x_{i}) -f( u) |\\ & < \frac{1}{2} \epsilon +\frac{1}{2} \epsilon =\epsilon \ \square \\ \end{aligned}\\ \end{gathered}

Notes

If UU is infinite, we are unable to take the minimum of all neighborhoods. To recap, our goal is to find a δ\delta that is “close enough” for arbitrary x, u Ax,\ u\ \in A such that f(x)f(u)<ϵ|f( x) -f( u) |< \epsilon. The idea is to use xix_{i} as a bridge between xx and uu since we only know how ff behaves near each xix_{i}. Choosing 12δxi\frac{1}{2} \delta _{ {x}_{i} }neighborhoods ensures x, ux,\ u are in the same neighborhood of xix_{i}. This enables us to rely on ff’s local behavior.

Bolzano-Weierstrass Theorem: Every bounded sequence in R\mathbb{R} has a convergent subsequence

The proof is only for R\mathbb{R}, but the theorem also holds in Rn\mathbb{R^n} and Cn\mathbb{C}^n.

Definition: Let (xn)(x_n) be a sequence of real numbers. xNx_N is called a peak if nN,xNxn\forall n\geq N, x_N\geq x_n i.e., xNx_N is the largest term in the sequence starting from it.

Lemma: Every sequence of real numbers has a monotone subsequence

Proof

Let (xn)(x_n) be a sequence of real numbers. There are two cases regrading the number of peaks:

Case1: (xn)(x_n) has infinitely many peaks. We can find a subsequence (xnk)(x_{n_{k} }) consisting of peaks. (xn1)(x_{n_{1} }) is a peak, so xn1xn,nn1x_{n_{1} } \geq x_n, \forall n\geq n_1. Thus, xn1xn2x_{n_{1} }\geq x_{n_{2} }. Similarly, xn2x_{n_{2} } is a peak, so xn2xn3xn4x_{n_{2} } \geq x_{n_{3} } \geq x_{n_{4} } \geq \cdots, etc.

The sequence is monotone increasing.

Case2: (xn)(x_n) has finitely many peaks. Then, we choose our subsequence after the last peak. Let xNx_N be the last peak. For xn,nN,xm,m>n,such that  xm>xnx_n, n\geq N, \exists x_m, m\gt n, \text{such that } \ x_m \gt x_n since xnx_n is not a peak. Similarly, xmx_m is not a peak, so xk,k>m,such that  xk>xm\exists x_k, k\gt m, \text{such that } \ x_k \gt x_m. Following this procedure, we end up with a monotone increasing subsequence.

In both cases, there exists a monotone subsequence in (xn)(x_n), so the lemma is proved.

Proof of the theorem

(xn)(x_n) is bounded. By lemma, (xn)(x_n) has a monotone subsequence (xnk)(x_{n_{k} }). By the monotone convergence theorem, the sequence converges \square.

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