My Favorite Proofs from Analysis1

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 $\mathbb{R}^2$)

Probably the most important inequality in this course.

$|x+y|\leq|x|+|y|$

Proof

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

For the triangle inequality:

Notes

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

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

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

Proof

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

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

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

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

Notes

The choice of $\frac{1}{n}<y-x$ in the first step seems mysterious, but everything follows that is easy to understand. The reason why we made this choice is: $\frac{1}{n}$ is the step size, if it’s greater than $y-x$, assume $z<x, z+\frac{1}{n}$ 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 $S =\{k\in\mathbb{N}:\frac{k}{n} \lt x\}$ may be empty ( $\frac{1}{n}\gt x$ ), but that doesn’t matter. $\mathbb{R}\setminus S$ is “more” infinite. The well-ordering principle holds as long as $\mathbb{R}\setminus S$ 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 $f: A→ \mathbb{R}$ be continuous. Let $\epsilon >0,$ then $\forall \ x\in \ A,\ \exists \delta _{x} >0:|x-u|< \delta _{x} \Longrightarrow \ |f( \ x) \ -f( \ u) |\ < \ \frac{1}{2} \epsilon$ (1)

Consider the neighborhoods $V*{\frac{1}{2} \delta *{x} }( x) ,\ \forall x\in A$. Then, $C\ :=\{V_{\frac{1}{2} \delta _{x} }( x) :x\in A\}$ is an open cover of A since an arbitrary union of open sets are still open. A is compact, so $C$ has a finite subcover, $U:=\{V*{\frac{1}{2} \delta *{x*{k} }}( x*{1}) ,\ \cdots ,V*{\frac{1}{2} \delta *{x*{k} }}( x*{k})\}$ where $x*{1} ,\cdots ,x*{k} \in A$

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

Intuition: If $x, u$ are in the same neighborhood around $x*{i}$, we can utilize the nice, local behavior. The problem is that $\delta *{x*{i} }$ depends on $x*{i}$, so the required distance varies in different locations on the domain. Can we find a $\delta$ 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 $\delta :=\min\left\{\frac{1}{2} \delta*{{x}*{1}},\cdots ,\frac{1}{2} \delta*{{x}*{k}}\right\}$. $U$ covers A, so $x \in V*{\frac{1}{2} \delta*{{x}_{i}}}( x*{i})$for some $x*{i} \in A$.

Notes

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

Bolzano-Weierstrass Theorem

Every bounded sequence in $\mathbb{R}$ has a convergent subsequence.

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

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

Lemma: Every sequence of real numbers has a monotone subsequence

Proof

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

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

The sequence is monotone increasing.

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

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

Proof of the theorem

$(x_n)$ is bounded. By lemma, $(x_n)$ has a monotone subsequence $(x_{n_{k} })$. 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