# Trigonometric Series and the Beginnings of Set Theory

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a $2\pi$-periodic function. It may or may not have a representation as a trigonometric series $\displaystyle{a_0+\sum_{n=1}^\infty a_n\sin(nx) + b_n\cos(nx)}$

A natural question to ask is whether or not the representation of $f$ as a trigonometric series is unique, if it has one. It was the consideration of this question that led Cantor to the invention of set theory.

There is a nice writeup of this story in the first part of this article by Alexander Kechris. I’ll give part of the story below.

Cantor solved the problem in the affirmative; i.e., he proved:

Suppose that a trigonometric series converges to zero everywhere in $\mathbb{R}$. Then all the coefficients of that series are zero.

(By subtraction, this is equivalent to the problem stated above.) He was also able to show (by a very similar method) the following, which I’ll call the Isolated Points Lemma:

Suppose $a < b < c$ and that a trigonometric series converges to zero on $(a,b)\cup (b,c)$. Then that series converges to zero at $b$ as well.

From these two results, we can immediately conclude the following:

Suppose that a trigonometric series converges to zero at all but finitely many points. Then the coefficients of that series are all zero.

Call a set $S\subset \mathbb{R}$ a set of uniqueness if whenever a trigonometric series converges to zero on $\mathbb{R} - S$, the coefficients of that series are all zero. Then the previous result may be stated: “All finite sets are sets of uniqueness.”

But we can use the Isolated Points Lemma to show more than that. For example, we can show that the set $S = \{1/n\mid n\in \mathbb{N}\}\cup \{0\}$ is a set of uniqueness. The reason is that if a trigonometric series converges to zero on $\mathbb{R} - S$, then by the Isolated Points lemma, it also converges to zero on the points in $\{1/n \mid n\in\mathbb{N}\}$.

But, now that we know that it converges to zero on the points in $\{1/n\mid n\in\mathbb{N}\}$, we can apply the Isolated Points Lemma again to show that it converges to zero at 0 (since we now know that at converges to zero on, e.g., $(-1,0)\cup (0,1)$).

What we have actually shown by the above argument is the following:

Given $A\subset\mathbb{R}$, let $A'$ be the set of limit points of $A$ (also known as the Cantor-Bendixson derivative of $A$). If $A'$ is a set of uniqueness, then $A$ is a set of uniqueness.

For any $A\subset \mathbb{R}$, let $A^{(n)}$ be the $n$th Cantor-Bendixson derivative of $A$. Then, by iterating the above fact, we have the following:

Suppose that for some $n$, $A^{(n)} = \emptyset$. Then $A$ is a set of uniqueness.

This is as far as we can go as long as we merely iterate the Cantor-Bendixson derivative finitely often. But, if we make the leap to iterating it transfinitely many times, we can go much further:

Theorem: All countable closed sets are sets of uniqueness.

Proof: First, define $A^{(\alpha)}$ for all ordinals and all closed sets $A$ as follows:

1. $A^{(0)} = A$
2. $A^{(\alpha + 1)} = (A^{(\alpha)})'$.
3. $A^{(\alpha)} = \bigcap_{\beta < \alpha} A^{(\beta)}$, when $\alpha$ is a limit ordinal.

The Isolated Points Lemma says that if a trigonometric series converges to zero outside of $A$, then it converges to zero outside of $A'$. We will generalize this by showing the following lemma:

Lemma: If a trigonometric series converges to zero outside of $A$, then it converges to zero outside of $A^{(\alpha)}$ for any $\alpha$.

Proof of Lemma: This is by transfinite induction. The successor step of the induction is just the Isolated Points Lemma again, so all we have to show is that, fixing a trigonometric series, if $\alpha$ is a limit ordinal and the series converges to zero outside of each $A^{(\beta)}$ for $\beta < \alpha$, then it converges to zero outside of $A^{(\alpha)}$. But this follows simply because every point of $\mathbb{R}-A^{(\alpha)}$ must be in some $\mathbb{R}-A^{(\beta)}$ for $\beta < \alpha$ by definition. End of proof of Lemma.

To complete the proof of the theorem then, we just have to observe that for all countable closed $A$, $A^{(\alpha)}=\emptyset$ for some $\alpha$. Clearly, for all closed $A$, there is an $\alpha$ such that $(A^{(\alpha)})'=A^{(\alpha)}$ (this is because $\{A^{(\beta)}\}$ is a decreasing sequence). But it is standard fact that a set $B$ such that $B' = B$ is either empty or of cardinality $2^{\aleph_0}$. (Such a set is called a perfect set and a reference for the cited fact is page 7 of David Marker’s notes on descriptive set theory.) End of proof.

As a historical note, Kechris reports that while thinking about the above issues led Cantor to discover ordinals, he never actually wrote down a proof of the above theorem; that was finally done by Lebesgue in 1903.

Further, it was later proven by Bernstein and Young independently that arbitrary countable sets are sets of uniqueness, and by Bari that countable unions of closed sets of uniqueness are sets of uniqueness.

Edit: Simplified the proof.

## 2 thoughts on “Trigonometric Series and the Beginnings of Set Theory”

1. tba says:

Hello,

Thank you for a very interesting post. Unfortunately my set theory is very poor so I got lost near the beginning:

Can someone explain how we know that if a series converges to 0 for all {1/n | n c N}, it must converge to 0 at 0? I don’t understand why we know it “converges to 0 on, e.g. (-1,0) U (1,0)”: how do we know that this is true on the infinitely many points of the form x/y (1<x<y,x is coprime to y)?

Thanks!

2. Christian says:

I think that the right formula for the Fourier Series includes $\frac{a_{0}}{2}$.