**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 $3=\{0,1,2\}$.



An infinite subset in $3^\omega$ with "large" projections?
**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.


