For any set $X$, let $[X]^2 = \big\{\{x,y\}:x\neq y\in X\big\}$.

Let $f:[\omega]^2\to\{0,1\}$ be a function. The principal goal is to find a partition of $\omega$ such that if $m\neq n\in \omega$ are in the same block of the partition, then $f(\{m,n\}) = 0$, and if $m,n$ are not in the same block, then $f(\{m,n\})= 1$.

It is easy to see that given $f:[\omega]^2\to\{0,1\}$ it is not always possible to find such a "perfect" partition that meets the requirement above.

So for any partition $P$ of $\omega$ we define the collection of wrong members of $[\omega]^2$, denoted by $W_P\subseteq [\omega]^2$ in the following way. Set

• $W_P^{(0)} = \big\{e\in[\omega]^2\setminus\big(\bigcup\{[b]^2:b \in P\}\big): f(e) = 0\big\}$, that is the set $W_P^{(0)}$ is the set of all $e\in[\omega]^2$ such that $e$ is not inside some block, but $f(e) = 0$ , and

• $W_P^{(1)} = \big\{e\in\bigcup\{[b]^2:b \in P\}: f(e) = 1\big\}$, that is the set $W_P^{(0)}$ is the set of all $e\in[\omega]^2$ such that $e$ is indeed inside some block, but $f(e) = 1$.

The intuition is that the members of $W_P^{(0)}$ are "falsely" labelled with $0$ by $f$, and the members of $W_P^{(1)}$ are "falsely" labelled with $1$ by $f$.

Question. Is there $f:[\omega]^2\to\{0,1\}$ with the following property?

Given any partition $P$ of $\omega$, is it always possible to find a partition $P_2$ of $\omega$ such that $W_{P_2}\subseteq W_P$ and $W_{P_2} \neq W_P$.

For any set $X$, let $[X]^2 = \big\{\{x,y\}:x\neq y\in X\big\}$.

Let $f:[\omega]^2\to\{0,1\}$ be a function. The principal goal is to find a partition of $\omega$ such that if $m\neq n\in \omega$ are in the same block of the partition, then $f(\{m,n\}) = 0$, and if $m,n$ are not in the same block, then $f(\{m,n\})= 1$.

It is easy to see that given $f:[\omega]^2\to\{0,1\}$ it is not always possible to find such a "perfect" partition that meets the requirement above.

So for any partition $P$ of $\omega$ we define the collection of wrong members of $[\omega]^2$, denoted by $W_P\subseteq [\omega]^2$ in the following way.

• $W_P^{(0)} = \big\{e\in[\omega]^2\setminus\big(\bigcup\{[b]^2:b \in P\}\big): f(e) = 0\big\}$, that is the set $W_P^{(0)}$ is the set of all $e\in[\omega]^2$ such that $e$ is not inside some block, but $f(e) = 0$ , and

• $W_P^{(1)} = \big\{e\in\bigcup\{[b]^2:b \in P\}: f(e) = 1\big\}$, that is the set $W_P^{(0)}$ is the set of all $e\in[\omega]^2$ such that $e$ is indeed inside some block, but $f(e) = 1$.

The intuition is that the members of $W_P^{(0)}$ are "falsely" labelled with $0$ by $f$, and the members of $W_P^{(1)}$ are "falsely" labelled with $1$ by $f$. Finally, we let $$W_P = W_P^{(0)} \cup W_P^{(1)}.$$

Question. Is there $f:[\omega]^2\to\{0,1\}$ with the following property?

Given any partition $P$ of $\omega$, is it always possible to find a partition $P_2$ of $\omega$ such that $W_{P_2}\subseteq W_P$ and $W_{P_2} \neq W_P$.