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What are the most prominent uses of transfinite induction in fields of mathematics other than set theory?

(Was it used in Cantor's investigations of trigonometric series?)

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    I find it the most natural and intuitive way to apply the axiom of choice in just about any context where it's needed. (more precisely, in the form of algorithms with transfinitely many steps) –  Jun 29 '15 at 19:24
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    It should be pointed out that many applications of Zorn's lemma can be reformulated as an application of transfinite induction. Instead of Zorn giving you a maximal element outright, you would start building it one step at a time and then argue that the construction cannot go on forever. – Miha Habič Jun 29 '15 at 19:33
  • Is your question specific about transfinite induction or about equivalent statements like Axiom of choice in general? I have rarely seen mathematical proves using transfinite induction. But more axiom of choice or even more frequently Zorn lemma. – mathcounterexamples.net Jun 29 '15 at 19:33
  • On Cantor's theorem about sets of uniqueness - See the proof of Theorem 4.2 and the remarks on pages 12-13 here: http://www.math.caltech.edu/~kechris/papers/uniqueness.pdf – hot_queen Jun 29 '15 at 20:23
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    Not a prominent example, but funny: there is a cover of Euclidean 3D space with non-intersecting circles with $r=1$. – CiaPan Jun 30 '15 at 07:20
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    @CiaPan could you point me to a reference for that statement? Sounds intriguing.. – Joachim Jun 30 '15 at 08:06
  • Any countable set of circles in a space lays in a union of at most countable set of planes. $\implies$ There exists another plane, different from all previous planes, so it intersects each circle in at most two points. $\implies$ The new plane contains at most countable set of points belonging to the circles. $\implies$ The plane has a place for another circle, disjoint with all previous circles. – CiaPan Jun 30 '15 at 08:22

10 Answers10

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Here's an example due to Erdos and Hajnal:

Theorem: There is a partition of plane into countably many pieces such that the distance between any two points in the same piece is irrational.

Corollary: Every non (Lebesgue) null subset $X$ of plane contains a non null subset $Y$ such that the distance between any two points of $Y$ is irrational.

Open question: Can we strengthen the corollary to: Every non null subset $X$ of plane contains a subset $Y$ of same outer measure as $X$ such that the distance between any two points of $Y$ is irrational?

Proof of theorem: By induction on $\kappa$, we show that

$(\star)$: Whenever $X \subseteq \mathbb{R}^2$ has size $\kappa$, there is a well orering $\preceq$ of $X$ such that for every $x \in X$, the set of $\preceq$-predecessors of $x$ which are at a rational distance from $X$ is finite.

Note that $(\star)$ suffices to construct a rational distance free partition of $X$ into countably many subsets.

When $\kappa \leq \omega$, this is obvious. So assume this is true for all $X \subseteq \mathbb{R}^2$ such that $|X| < \kappa$ where $\kappa \geq \aleph_1$. Let $|X| = \kappa$. Inductively construct $\langle X_i : i < \kappa \rangle$ such that the following hold:

(0) $X_i$'s are increasing continuous and $X = \bigcup \{X_i : i < \kappa\}$

(1) $|X_i| = \max(\aleph_0, |i|)$

(2) Whenever $x \neq y$ are from $X_i$, $z \in X$ and the distance of $z$ from each one of $x, y$ is rational, $z \in X_i$

Let $\preceq_i$ be a well order on $X_i$ witnessing $(\star)$. Define a well order $\preceq$ on $X$ as follows: If $x, y \in X_i \setminus \bigcup \{X_j : j < i\}$, then $x \preceq y$ iff $x \preceq_i y$. If $x \in X_i \setminus \bigcup \{X_j : j < i\}, y \in \bigcup \{X_j : j < i\}$, then $y \preceq x$. It is easy to check that $\preceq$ witnesses $(\star)$.

Komjath extended this to $\mathbb{R}^n$ for every $n$. The proof is slightly more complicated - $(\star)$ is replaced by a different statement which is again proved by transfinite induction (Note that $(\star)$ is false when $n \geq 3$).

hot_queen
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    What is $\kappa$? Is it an ordinal? ${}\qquad{}$ – Michael Hardy Jun 29 '15 at 21:00
  • The corollary is misstated. It is trivial if $Y$ is allowed to be a singleton set, and it is false if $Y$ is required not to be singleton. – Colin McLarty Jun 30 '15 at 06:34
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    The open question seems to have been settled affirmatively in 2012, for Lebesgue measure . http://arxiv.org/pdf/1207.5029.pdf – Colin McLarty Jun 30 '15 at 08:36
  • @ColinMcLarty I think the corollary is fine. Also the paper you quote only resolves the problem in dimension 1. – hot_queen Jul 01 '15 at 01:10
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    @hot_queen No matter what $X$ is, every singleton subset $Y$ of $X$ satisfies the condition trivially since there are no two different points in $Y$ right? That does not use Erdos and Hajnal's result at all. So that cannot be what you meant to call a corollary to their result. – Colin McLarty Jul 01 '15 at 06:02
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    @ColinMcLarty Maybe you are confusing non null with non empty? By non null I mean not of zero area. – hot_queen Jul 01 '15 at 12:31
  • @hot_queen Yes. sorry. – Colin McLarty Jul 02 '15 at 05:08
  • @hot_queen: When looking at your MO answers it's obvious you're an expert in set theory.I would like to ask you if the example by Erdős which I've added below is a well known example when talking about transfinite induction. – Markus Scheuer Jul 05 '15 at 11:19
  • I believe it is very well known and widely quoted result. For example, it appears in "Proofs from the book". It is also mentioned (without proof) in Kunen's article in Handbook of mathematical logic. – hot_queen Jul 05 '15 at 12:45
  • Hey Markus, Your reference to Erdos' paper led me to a problem he asked at the end of that paper. I could't see how to do it so I asked on mathoverflow. I think/hope the solution should be interesting. – hot_queen Jul 05 '15 at 21:13
  • Thanks for your answer and you're right, Erdös proof is even in Proofs from the book. So, it's definitely well known. :-) I'll have a look at MO to see your question, thanks again. Btw. if you start a comment with @: the addressee will receive a notification. Best regards, – Markus Scheuer Jul 12 '15 at 08:33
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Here are two contributions, one regarding complex analysis, the other number theory. But first some information to OPs question regarding Cantor and transfinite induction.

Historical Notes:

The following is mainly based upon The Search for Mathematical Roots 1870-1940 by I. Grattan-Guinness. The author mentions in chapter 3: Cantor: Mathematics as Mengenlehre, that Cantor's remainder came into his hands in the late 1960s. Cantor's family asked him to set the papers in some order and then they were placed to the University of Göttingen.

  • Über die Ausdehnung eines Satzes der Theorie der trigonometrischen Reihen (1872): This paper from 1872 shows his results regarding trigonometric series. He introduced the fundamental concept of limit-point (Grenzpunkt) of a set $P$. Then Cantor defined derived point-sets (abgeleitete Punktmengen) of $P$, each one comprising the set of limit-points of its predecessor. But he did not develop a line of thought regarding infinitieth derived sets $P^{(\infty)}$ at this stage.

According to I. Grattan-Guinnes Cantor's theory was doubtless a bit too intuitive at that stage. So, the concepts of well-ordered sets or transfinite induction were out of reach at that time.

  • Über unendliche, lineare Punktmannichfaltigkeiten (1879): Beginning with an extended study of the derivation of point-sets.

  • Beiträge zur Begründung der transfiniten Mengenlehre (1897): This is the paper where ordinal numbers, limit-ordinals and well-ordered sets were defined and the principle of transfinite induction was used the first time.

Theorem by Paul Erdős

The following theorem by Paul Erdős was proved using transfinite induction.

Suppose, we have a family $\{f_\alpha\}$ of holomorphic functions such that for each $z$ the set of values $f_\alpha(z)$ is countable. Let us denote this property $P_0$. Does it then follow that the family itself is countable?

The asthonishing answer is: it depends, namely on the continuum hypothesis.

Theorem (Erdős): If $c>{ \aleph_1}$, then every family $\{f_\alpha\}$ with property $P_0$ is denumerable. if $c=\aleph_1$, some family $\{f_\alpha\}$ with property $P_0$ has the cardinality $c$.

The original, short proof by Erdős can be found in a paper from 1963. The transfinite induction is an essential part of the proof in the case $c=\aleph{_1}$.

$$$$

Goodstein's Theorem

The following theorem about Goodstein sequences seems to be at the first glance purely number-theoretic, but it has some more interesting consequences.

Let $m,n$ be natural numbers $n> 1$. Write numbers $m$ as base $n$ representation of $m$ as follows. First write $m$ as sum of powers of $n$. For example, if $m=266$ and $n=2$ we obtain $$266=2^8+2^3+2^1$$ Then write each exponent as the sum of powers of $n$ and repeat this process until the representation stabilizes. $$266=2^{2^{2+1}}+2^{2+1}+2^1$$ Now define the number $G_n(m)$ as follows. If $m=0$ set $G_n(m)=0$. Otherwise set $G_n(m)$ to be the number produced by replacing every $n$ in the base $n$ representation of $m$ by $n+1$ and then subtracting $1$.

The Goodstein sequence for $m$ is defined by $$m_0=m, m_1=G_2(m_0), m_2=G_3(m_1), m_3=G_4(m_2),\ldots$$

So, for example, \begin{align*} 266_0&=266=2^{2^{2+1}}+2^{2+1}+2\\ 266_1&=3^{3^{3+1}}+3^{3+1}+2 \sim 10^{38}\\ 266_2&=4^{4^{4+1}}+4^{4+1}+1\sim 10^{616}\\ 266_3&=5^{5^{5+1}}+5^{5+1}\sim10^{10000} \end{align*}

Now, despite the fact that some Goodstein sequences increase enormously fast at the beginning, the Goodstein's Theorem surprisingly states that a Goodstein sequence for $m$ starting at $n$ eventually hits zero.

The proof is done using transfinite induction and the theorem is historically significant, since it was one of the first presenting a true statement which is unprovable in Peano Arithmetic. This was shown in Accessible independence results for Peano arithmetic by Laurie Kirby and Jeff Paris in 1982. (Definition and example above are from this paper).

Here is a related MO answer.

Markus Scheuer
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Showing that there are exactly continuum many Borel subsets of the real line uses transfinite induction on some measure of complexity of the set relative to the open sets.

Miha Habič
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  • I wonder if this would be "descriptive set theory" and therefore excluded by the phrasing of the question? – Michael Hardy Jun 29 '15 at 19:31
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    @MichaelHardy Perhaps, but I feel the result is basic enough for it to be equally well placed within topology or measure theory. – Miha Habič Jun 29 '15 at 19:34
  • . . . . . and now I'm thinking there must be other ways to do this. But that's not saying much, since there are always other ways. – Michael Hardy Dec 17 '17 at 00:09
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I think that Zorn's Lemma made it possible for mathematicians who are not familiar with set theory to use transfinite induction. For example, every commutative ring has a maximal ideal, you can prove that with transfinite induction, or use Zorn's Lemma.

Learning transfinite induction, and the theory of ordinals, is a bit time consuming. Many mathematicians are perfectly fine by learning this simple one lemma, and avoiding to learn all else.

Someone correct me if I am wrong, but before Zorn's Lemma mathematicians based their arguments on the transfinite. Since then hardly anybody does this anymore.

  • One thing you sort of said but which deserves to be emphasized: using Zorn's Lemma is using transfinite induction. The proof of Zorn's Lemma uses transfinite induction, and the statement is really just a "template" for a very common kind of transfinite induction argument, where the exact hypotheses needed for the argument have been abstracted away. – Eric Wofsey Jul 04 '15 at 22:48
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By understanding transfinite induction by one of your answers earlier today, to this Question that I asked, I am also trying to understand the following result.

Let $\mathcal{E}$ be the class of elementary amenable groups, then using transfinite induction, we can give another description for the class of groups $\mathcal{E}$. Let $\mathcal{A}$ be the class of finitely generated abelian-by-finite groups and for each ordinal $\alpha$, define $\mathcal{L_\alpha}$ inductively as follows-

\begin{align} \mathcal{L_0} & =\{1\} \\ \mathcal{L_\alpha}& =(L\mathcal{L_{\alpha-1}})\mathcal{A} & & \text{if } \alpha \text{ is a successor ordinal},\\ \mathcal{L_\alpha} & =\bigcup_{\beta < \alpha}\mathcal{L_\beta} & & \text{if } \alpha \text{ is a limit ordinal} \end{align}

Setting $\mathcal{L}=\bigcup_{\alpha\ge 0}\mathcal{L}_\alpha$, we can now state the following results.

$(\text{i} )\ \mathcal{L}=\mathcal{E}\\ \text{(ii)}\ \text{Each}\ \mathcal{L}_\alpha\ \text{is subgroup closed}$

This allows us to associate each $G\in \mathcal{E}$ with the least ordinal $\alpha$ such that $G\in \mathcal{L}_\alpha$.

In his paper Applications of a New $K$-Theoretic Theorem to Soluble Group Rings , Kropholler uses it as a lemma, along with many other results to settle Kaplansky's Zero Divisor Conjecture for soluble groups.

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A promiment theorem that comes to mind is

Every vector space has a basis. More precisely, for every vector space $V$ over scalar field $F$ there is a (possibly very large!) basiss $B \subset F$ such that every vector $\mathbf{v} \in V$ can be represented uniquely as a finite sum $\mathbf{v} = \sum_{i=1}^k x_k \mathbf{b}_k$ with $x_1,\ldots,x_k \in F \setminus \{0\}$ and $\mathbf{b}_1,\ldots,\mathbf{b}_k \in B$. (Note $k$ will of course depend on $\mathbf{v}$).

This is shown by (in a sense) constructing $B$ explicitly. You simply add elements to $B$ until every $\mathbf{v}\in V$ can be represented, while taking care not to add any elements which would allow a vector to be represented by two different sums.

As long as the cardinality of $V$ is finite or countable, this is straight-forward induction, but to extend the theorem to arbitrary vector spaces, you need transfinite induction to formalize the proof. The easiest way of doing so is probably by using Zorn's lemma.

fgp
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    I found the well-ordering theorem the easiest, since the main obstacle to formalizing the algorithm is a way to pick an element at each step. Zorn's lemma feels like a different approach; sort of like how people resort to $\epsilon$-$\delta$ proofs when they're really thinking in terms of infinitesimals. –  Jun 30 '15 at 14:12
  • @Hurkyl: That's not the only difference. When using transfinite induction one is 'counting' the elements of some set. When using Zorn's lemma one has to make sure that one does not accidentally ask for a set that does not exist because it is too large, like "the set of all field extensions of F". – user21820 Aug 05 '15 at 09:43
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The small object argument is a very important technique based on transfinite induction allowing for the construction of approximations/factorizations in various algebraic and topological contexts.

Hanno
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The characterization of all real functions such that $f(x+y)=f(x)+f(y)$ for all $x,y\in\Bbb R$ relies on transfinite induction on a well-ordering of the real numbers. That question with its variations pops up here at this site from time to time.

Rory Daulton
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    Don't you need "merely" a (Hamel) basis of $\mathbb R$ over $\mathbb Q$? – Hagen von Eitzen Jun 30 '15 at 12:07
  • @HagenvonEitzen: Yes, but the proof that I have seen of the existence of a Hamel basis uses transfinite induction on the well-ordering of the reals, or equivalently the Axiom of Choice. Also, people asking this question do not know of Hamel bases and need an answer from more basic principles. I did not know the name of the Hamel basis until I read your comment! – Rory Daulton Jun 30 '15 at 16:03
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A historically significant example is that of the Cantor-Bendixson rank which is in fact the result that motivated Cantor to invent transfinite induction, coming from the problem of describing the set where a Fourier series can converge/diverge.

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There are several proofs of basic theorems of general topology which are proved by transfinite induction —for example, metrization theorems. You can find the very beautiful arguments explained in Hu's book on general topology.