There is an infinite sequence of functions $g_1,g_2,\ldots$, each of them is $\Bbb R\to\Bbb R$. Prove that there exists a finite set of functions $f_1,f_2,\ldots,f_n$ such that any function $g_k$ can be expressed as a composition $f_{k_1}\circ f_{k_2}\circ\cdots\circ f_{k_m}$.


The main ingredient in my answer is the following fact:

$|\mathbb{R}^{\mathbb N}| = |\mathbb{R}|$.

Its proof isn't directly related to this question so I won't put it right here. But you can find it in the answer to this question. Due to this fact we can talk about the $\mathbb R$ as about the disjoint union $\coprod\limits_{i=1}^\infty R_i$ where all of the $R_i$'s are the copies of $\mathbb{R}$.

Now we can define two functions $\mathbb{R} \to \mathbb{R}$: $F$ and $G$. $F$ sends $R_i$ due to the map $g_i$ (namely, $F|_{R_i} = g_i \circ T_i$, where $T_i$ is an arbitrary bijection from $R_i$ to $\mathbb R$) and $G$ works by the following rule: it's defined as a function $\coprod\limits_{i=1}^\infty R_i \to \coprod\limits_{i=1}^\infty R_i$ which sends $T_{i-1}^{-1}(a)$ to $T_i^{-1}(a)$ for all $a$ in $\mathbb R$. It's a well-defined map since all $T_i$'s are bijections.

The rest of the proof is quite elementary: you can easily check that $g_i = F \circ G^{i-1} \circ {T_1}^{-1}$. So your finite set of functions $f_i$ is just $\{F, G, T_1\}$.

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    $\begingroup$ By $|\mathbb R|^{|\mathbb N|}=|\mathbb R|$ don't you mean $|\mathbb N|\cdot|\mathbb R|=|\mathbb R|$? $\endgroup$ – bof Sep 14 at 19:24
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    $\begingroup$ Also, the finite set should have $T_1^{-1}$, rather than $T_1$. $\endgroup$ – Eric Wofsey Sep 15 at 1:19
  • $\begingroup$ what does the syntax $F|_{R_i}$ represent? If it means $F$ for an input from $R_i$ then shouldn't $F$ be a set of functions? $\endgroup$ – user3338098 Sep 19 at 19:42

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.)

Since you already accepted an answer, I won't bother to give a proof, but I will give references. The theorem is due to Sierpiński:

W. Sierpiński, Sur les suites infinies de fonctions définies dans les ensembles quelconques, Fund. Math. 25 (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) 230–252.)

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.


Let $b:{\Bbb R}\to [0,1)$ be a bijection and $u:x\mapsto x+1$ be a unit shift. We can define $G:[1,\infty)\to{\Bbb R}$ as $$ G(x)=g_{[x]}\left(b^{-1}\left(\{x\}\right)\right) $$ where $[x]$ and $\{x\}=x-[x]$ are integer and fractional parts of $x$ respectively. Now we can write the $n$-th function in the original sequence as

$$ g_n(x) = G\left(n+b(x)\right)\quad\mbox{for $n\in{\Bbb N}$.} $$

In other words, each $g$ can be expressed as a composition of $f_1=b$, $f_2=G$ and some number of $f_3=u$'s:

\begin{equation} \qquad g_n = G\circ u^{(n)}\circ b\qquad ∎ \end{equation}


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