2
$\begingroup$

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 $x\big|_J=y\big|_J\Rightarrow f(x)=f(y)$.

  • 0
    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.)2012-02-22

1 Answers 1

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.

  • 0
    You’re assuming that the OP’s $I$ is the unit interval. What if it’s a compact space of cardinality greater than $2^\omega$?2012-02-26