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Let $I$ a compact set and $f\in\mathcal{C}([0,1]^I)$. Then exists $J\subset I$, countable or finite, such that:

if $x,y\in [0,1]^I$ such that $x\big|_J=y\big|_J\Rightarrow f(x)=f(y)$.

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Technical note: the domain is countable-dimensional, not countable. –  Alex Becker Feb 22 '12 at 22:15
    
Clearly, if $J$ is a dense subset of $I$, then $x|_J=y|_J$ implies $x=y$, so $f(x)=f(y)$. Therefore, if not true, there cannot be a countable dense subset of $I$. Not sure if that helps. (That assumes that $[0,1]^I$ means the set of continuous functions from $I$ to $[0,1]$, not the set of all functions.) –  Thomas Andrews Feb 22 '12 at 22:22

1 Answer 1

The result is true for the set of all functions from $I$ into $[0,1]$ (not only the continuous ones).

Claim. For any open subset $U\subset [0,1]^I$ there exists $J\subset I$ countable so that $x\in \overline U$ and $x|_J=y|_J$ imply $y\in \overline U$.

Find a maximal pairwise disjoint family of standard basic nonempty open subsets of $U$, by the Hewitt-Marczewski-Pondiczery theorem this family is countable, let us call it $(U_n)$. For each $n$, let $J_n$ be the set of non-trivial coordinates in the basic open set $U_n$. It should be clear that $J=\bigcup_n J_n$ satisfies the requirements of the claim.

Let $\{ V_n:n\in\mathbb N\}$ be a countable base of $\mathbb R$. For each n, we can find a countable subset $J_n$ of $I$ satisfying the conclusion of the Claim with respect to the open set $f^{-1}(V_n)$. Now, we claim that $J=\bigcup J_n$ is the required set. Suppose $x|_J=y|_J$ and $f(x)\ne f(y)$ then there are disjoint open sets $V_n, V_m$ so that $f(x)\in V_n$ and $f(y)\in V_m$. This implies that $x\in f^{-1}(V_n)=U_n$ and $f^{-1}(V_m)=U_m$, the sets $U_n, U_m$ are disjoint. But by the claim $y\in U_m$ and $x|_{J_m}=y|_{J_m}$ implies $x\in \overline{U_m}$ which is impossible.

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Where have you used that $I$ is compact? If $[0,1]^I$ is all functions, then the only topology I know on that set is essentially the $I$-index product topology on $[0,1]$, which depends only on the cardinality of $I$ and not the topology of $I$. –  Thomas Andrews Feb 23 '12 at 3:53
    
@ThomasAndrews. I didn't use the compactness. (It is not necessary). I suppose as you mentioned it before that the OP had in mind the set of continuous functions from $I$ into $[0,1]$. So there is, probably, an easier proof in that case. –  azarel Feb 23 '12 at 5:58
    
What is... "maximal pairwise disjoint family of standard basic nonempty open subsets"? –  Nerea Feb 23 '12 at 11:38
    
You’re assuming that the OP’s $I$ is the unit interval. What if it’s a compact space of cardinality greater than $2^\omega$? –  Brian M. Scott Feb 26 '12 at 12:35

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