Problem. Is there an infinite set $I\subset 3^\omega$ such that for any infinite subset $J\subset I$ there exists $n\in\omega$ such that $\{x(n):x\in J\}=3$?

Here $3^\omega$ is the set of functions from $\omega$ to $3=\{0,1,2\}$.


1 Answer 1


Claim: For every integer $k>1$ there exists a subset $I\subseteq k^\omega$ of cardinality $\frak{c}$ such that for every subset $J\subseteq I$ such that $|J|=k$ there exists $n\in\omega$ such that $|\{x(n)\mid x\in J\}|=k$.

Proof: By Theorem 7.7 of Jech "Set Theory" there exists an independent family $\mathcal{F}$ of subsets of $\omega$ of cardinality $\frak{c}$. Independent means that for every distinct members of $\mathcal{F}$ $F_1,F_2,\ldots,F_s,F_1',F_2',\ldots,F'_t$ the intersection $$ F_1\cap F_2\cap \ldots\cap F_s\cap (\omega\setminus F_1')\cap (\omega\setminus F_2')\cap \ldots\cap (\omega\setminus F'_t) $$ is nonempty. Partition $\mathcal{F}$ into subsets of size $k-1$: $S_i=\{F_{i1},F_{i2},\ldots,F_{ik-1}\}$. We have $\frak{c}$ such subsets.

Define the subset $I=\{x_i\}$ as: $$ x_i(n) = |\{F\in S_i\mid n\in F\}| $$

Let $J\subseteq I$ be a subset of size $k$. Let $S_0,S_1,\ldots,S_{k-1}$ be the subsets of $\mathcal{F}$ that correspond to the elements of $J$. Since the family $S_0\cup S_1\cup \ldots\cup S_{k-1}$ is independent we can find an $n\in\omega$ which belongs to no members of $S_0$, to one member of $S_1$, and so on.


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