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Previously asked and bountied at MSE with slightly different language:

Given a topological space $\mathcal{X}$, let $$\mathsf{Cl_C}(\mathcal{X})=\bigcup_{n\in\mathbb{N}}C(\mathcal{X}^n,\mathcal{X})$$ be the clone of finite-arity continuous functions on $\mathcal{X}$ (where $\mathcal{X}^n$ is given the product topology). Motivated by the Kolmogorov-Arnold superposition theorem, let the superposition number of $\mathcal{X}$ be the minimal cardinality of a set of continuous functions $\mathfrak{F}\subseteq\mathsf{Cl_C}(\mathcal{X})$ such that the clone generated by $\mathfrak{F}\cup C(\mathcal{X},\mathcal{X})$ is all of $\mathsf{Cl_C}(\mathcal{X})$.

Some quick observations:

  • Any space homeomorphic to its square trivially has superposition number $1$: there is no real distinction between single- and multi-variable continuous functions over such a space once we bring in an appropriate pairing function.

  • The K/A theorem says that the superposition number of $[0,1]$ is $1$, witnessed by $\{+\}$.

  • It is not clear to me that shifting attention to maps with codomain $[0,1]$ results in the same notion (I strongly suspect it doesn't), so e.g. Ostrand's extension of the K/A theorem doesn't seem relevant here.

I'm generally curious for any relevant information, but the following particular question seems interesting: is there, for each finite $n$, a "nice" topological space with superposition number $n$? Of course this is really a whole family of questions, one for each meaning of "nice." Tentatively I think the most interesting case is likely to be the following:

Question: Is there, for each finite $n$, a connected topological manifold (with or without boundary, not necessarily compact) with superposition number $n$?

I suspect that every topological manifold has superposition number $1$, by some not-too-complicated extension of K/A, but I don't immediately see how to show that.

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  • $\begingroup$ What do you mean with a "clone"? $\endgroup$
    – Wojowu
    Commented Dec 23, 2021 at 21:13
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    $\begingroup$ @Wojowu Clone in the sense of universal algebra. To get the clone generated by $\mathfrak{G}$, you add in all the projection functions and the close under composition. The set of finite-arity continuous functions constitutes a clone, but $\mathfrak{F}\cup C(\mathcal{X},\mathcal{X})$ will (usually) merely generate a clone. (I've added a link to the definition to my question.) $\endgroup$ Commented Dec 23, 2021 at 21:25

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