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Is there a reasonably well-behaved topological space $X$ (ideally Polish), a set $\kappa$, and a continuous function $g: X^\kappa\to\mathbb{R}$ that depends on uncountable many coordinates?

If $X$ is a compact Hausdorff space, the answer is known to be no. To see this, note that the family of all continuous functions depending on only finitely many coordinates satisfies the conditions for the Stone-Weierstrass theorem and is, therefore, uniformly dense. The argument can be found in textbooks.

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    $\begingroup$ It seems like a negative answer follows pretty directly from the definition of the product topology and the fact that $\mathbb{R}$ is second countable. Is there a subtlety that I'm missing? $\endgroup$
    – Nate Eldredge
    Commented Jun 24, 2020 at 21:58
  • $\begingroup$ I don't think so. Thank you! $\endgroup$
    – Michael Greinecker
    Commented Jun 24, 2020 at 22:13
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    $\begingroup$ As I found by writing a wrong answer, it is not so simple as that. The argument I had in mind shows that for each given $x$, there are countably many coordinates such that if $y$ agrees with $x$ in those coordinates, then $g(y)=g(x)$. But that's not good enough. $\endgroup$
    – Nate Eldredge
    Commented Jun 25, 2020 at 14:27
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    $\begingroup$ Do you know whether $X^\kappa$ is Lindelöf, at least for small $\kappa$? That would give a negative answer as well. $\endgroup$
    – Jonathan Schilhan
    Commented Jun 25, 2020 at 16:14

2 Answers 2

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Bockstein's theorem

Bockstein, M., Un théorème de séparabilité pour les produits topologiques, Fundam. Math. 35, 242-246 (1948). ZBL0032.19103.

This is the case of a product $\prod_{t \in T} X_t$ where all factors are second-countable. I that case any continuous function $\prod_{t \in T} X_t \to \mathbb R$ depends on countably many coordinates.

PLUG... See Theorem 2.1 in

Edgar, G. A., Measurability in a Banach space, Indiana Univ. Math. J. 26, 663-677 (1977). ZBL0361.46017.

where the special case $X = \mathbb R$ is done. That is, a continuous function $\mathbb R^T \to \mathbb R$ depends on only countably many coordinates.

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    $\begingroup$ Thank you! An English language proof of Bockstein's theorem with an application to my question can be found in: Ross and Stone. Products of Separable Spaces. 1964. $\endgroup$
    – Michael Greinecker
    Commented Jun 25, 2020 at 16:49
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    $\begingroup$ They also show that second-countable can be weakened to separable. $\endgroup$
    – Michael Greinecker
    Commented Jun 25, 2020 at 18:07
  • $\begingroup$ "They" = Ross & Stone, not Bockstein? $\endgroup$
    – Gerald Edgar
    Commented Jun 25, 2020 at 19:19
  • $\begingroup$ I can't read French; I assumed from your description that Bockstein's argument required the space to be second countable. $\endgroup$
    – Michael Greinecker
    Commented Jun 25, 2020 at 19:22
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Let $X$ be an uncountable discrete space with a distinguished element $0$. We view the product space $X^X$ as the space of maps $\phi: X \to X$. The set $$ E := \{ \phi \in X^X: \phi(\phi(0)) = 0 \}$$ is easily seen to be clopen, hence the indicator function $1_E: X^X \to {\bf R}$ is continuous, but depends on all of the (uncountably many) coordinates of $X^X$.

The key point here (which was inspired by Nate's comment based on the earlier incorrect attempt at solving this problem) is that deciding whether a given map $\phi$ belongs to $E$ requires only a finite number of (adaptive) evaluations of $\phi$, but the set of (non-adaptive) locations where $\phi$ could potentially need to be evaluated is uncountable.

Note that a similar construction works for $X \times \{0,1\}^X$ using the set $E := \{ (x, \phi) \in X \times \{0,1\}^X: \phi(x)=0\}$; thus even a single highly non-compact factor is enough to generate a counterexample. (But I am not sure what happens if one insists that all of the factors be sigma-compact, in particular can one construct a continuous function $f: {\bf N}^{\bf R} \to {\bf R}$ that depends on uncountably many coordinates?)

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  • $\begingroup$ Actually the case $X$ separable boils down to $X=\mathbf{N}$ which you suggest (just considering a map with dense image $\mathbf{N}\to X$ and composing with $\mathbf{N}^\kappa\to\mathbf{X}^\kappa$. $\endgroup$
    – YCor
    Commented Jun 25, 2020 at 16:19
  • $\begingroup$ This is a beautiful example !!! (but I think there is a typo, the indicator function 1_E should be defined on X^X and not on X. I tried to change that, but for some reason edits need to change at least 6 characters and I did not know what to change in this wonderdul answer... $\endgroup$
    – Cosine
    Commented Aug 29, 2023 at 9:05
  • $\begingroup$ Typo corrected, thanks. $\endgroup$
    – Terry Tao
    Commented Aug 30, 2023 at 11:36

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