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Given an infinite sequence of functions $\{g_1, g_2, \ldots, g_n, \ldots\}$ where $ g_n : \Bbb R \to \Bbb R$ prove there's a finite set of functions $ \{ f_1, f_2, \ldots, f_M \} $ such that any $ g_n $ can be represented as a composition of $ f_m $'s.

Honestly, not sure even how to approach this. The intuition is that if the infinite sequence of functions is not defined using finite set of functions and composition then the sequence definition would be infinite itself, but I don't know how to formalize that.

Batominovski
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Daniil Kolesnichenko
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  • Just to be sure: Is $R$ the set of real numbers? – Aryaman Maithani Jul 08 '20 at 11:32
  • @AryamanMaithani yes – Daniil Kolesnichenko Jul 08 '20 at 11:35
  • Interesting question! Have you tried any examples yourself? For example, the case where $g_n(x) = x^n$? Maybe you could also give some context as to where you found this. – Aryaman Maithani Jul 08 '20 at 11:38
  • Is this sequence defined in such generality? Do $g_n$ have any other property? Let's say, they continuous? Or maybe there is some kind of relation between each $g_n$? – alphaomega Jul 08 '20 at 11:47
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    @AryamanMaithani I think I see what you are suggesting — I probably didn't write the condition correctly. There wasn't a $ f_m : R \to R $ constraint in the original exercise, and when I tried the example you given I saw that it makes sense: it could be represented as $ t(x) = (x, 1); p(x, y) = (x, y * x); s(x, y) = y; g_1 = s \circ p \circ t; g_2 = s \circ p \circ p \circ t $ etc but I don't see a way to do that without using tuples (i.e. $R \to R \times R$ etc functions). I'm sorry, that's my bad, I'll edit the question. – Daniil Kolesnichenko Jul 08 '20 at 11:55

2 Answers2

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Fix a bijection $\varphi:\mathbb{R}\to [0,1)$ and define $\psi : \mathbb{R} \to \mathbb{R}$ by

$$ \psi(x) = \begin{cases} g_n(\varphi^{-1}(x-n)), & \text{if $x \in [n, n+1)$ for some $n \in \mathbb{N}_1$}; \\ 0, &\text{otherwise}; \end{cases} $$

where $\mathbb{N}_1 = \{1,2,3,\dots\}$. Finally, set $ f(x) = x+1 $. Then

$$ g_n = \psi \circ f^{\circ n} \circ \varphi $$

for any $n \in \mathbb{N}_1$.

Sangchul Lee
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    +1, very nice! (To elaborate: you basically use that $\Bbb R$ is a countable union of sets with the same cardinality as $\Bbb R$. So you can "encode" all the $g_n$s into one function $\psi$ and then retrieve them back.) – Aryaman Maithani Jul 08 '20 at 12:06
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    So... my intuition lead me in a completely wrong direction :)

    Thank you and thanks to @AryamanMaithani for elaboration, seems like a nice solution to me too

     – Daniil Kolesnichenko Jul 08 '20 at 12:16
  • I am most probably missing something, but what is the finite set of ${f_1, \cdots , f_M}$? we still need infinite number of functions, namely all the functions $f^{\circ n}$, where $n\in \mathbb{N}$. – alphaomega Jul 08 '20 at 12:20
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    @alphaomega, My interpretation was that we are allowed to use the same function as many times as we wish. In other words, we allow to write $$g_n=f_{i_1}\circ\dots\circ f_{i_N}$$ for some $N\in\mathbb{N}1$ and $i_1,\dots,i_N\in{1,\dots,M}$ but we do not require that $i_k$'s are distinct. (If $i_k$'s are required to be distinct, then there are at most $\sum{k=1}^{M}\binom{M}{k}k!$ possible combinations of compositions, which is of course finite.) – Sangchul Lee Jul 08 '20 at 12:23
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    @alphaomega it's just n compositions of f, so $ g_1 = \psi \circ f \circ \phi; g_2 = \psi \circ f \circ f \circ \phi $ etc, so we just need these three functions – Daniil Kolesnichenko Jul 08 '20 at 12:23
  • as usual, very nice answer ! – alphaomega Jul 08 '20 at 12:28
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Copying my answer from another question, which is closed and threatened with deletion:

More generally:

Theorem. For any infinite set $S$ and any countable set $F$ of functions $f:S\to S$, there are two functions $g,h:S\to S$ such that $F$ is contained in the semigroup generated by $\{g,h\}$ under composition. (On the other hand, if $S$ is a finite set with more than two elements, then three selfmaps of $S$ are needed in order to generate them all.)

Proof. Let $F=\{f_n:n\in\omega\}$. We may assume that $S=X\times\omega\times2$ for some infinite set $X$. Construct a bijection $g:S\to\{(x,n,i)\in S:n=0\text{ or }i=1\}$ such that $g(x,n,0)=(x,n-1,1)$ for $n\gt0$; thus $g^2[S]=X\times\{0\}\times\{0\}$. Define $h:S\to S$ so that $h(x,n,0)=(x,n+1,0)$ and $h(x,n,1)=f_ng^{-2}(x,0,0)$. Then $f_n=hgh^{n+1}g^2$.

The theorem is due to Sierpiński:

W. Sierpiński, Sur les suites infinies de fonctions définies dans les ensembles quelconques, Fund. Math. 24 (1935) 209–212.

A simpler proof of Sierpiński's theorem was given by Banach:

Stefan Banach, Sur un théorème de M. Sierpiński, Fund Math. 25 (1935) 5–6.

(By the way, if the given functions are bijections, then the functions $g,h$ can also be taken to be bijections; see Theorem 3.5 of Fred Galvin, Generating countable sets of permutations, J. London Math. Soc. (2) 51 (1995) 230–242.)

Sierpiński's theorem resurfaced in Monthly problem 6244, proposed by John Myhill; the solution appeared in Amer. Math. Monthly 87 (1980) 676–678.


Since every semigroup is isomorphic to a semigroup of mappings, as a corollary to Sierpiński's theorem we have:

Corollary. Every countable semigroup is embeddable in a $2$-generator semigroup.

This was proved in a different way by Trevor Evans, Embedding theorems for multiplicative systems and projective geometries, Proc. Amer. Math. Soc. 3 (1952) 614–620.

bof
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