does this implication relation induced by a subset of the power set have a name?

Question

Let $S$ be a nonempty set and $F\subseteq\mathcal{P}(S)$. Define $x\Rightarrow_F\!y$ if and only if $$\{A\in F:x\in A\}\subseteq\{A\in F:y\in A\}$$ and $x\!\!\iff_{\!\!\!\!F}\,y$ if and only if $x\Rightarrow_F\!y$ and $y\Rightarrow_F\!x$. Does the partial order $\leq_F$ on the quotient $S\big/\lower1pt\hbox{$\!\!\iff_{\!\!\!\!F}$}$ defined by x $\leq_F$ y if and only if $x\Rightarrow_F\!y$ for some $x\in$ x and $y\in$ y have a name?

Answer 1

I don't know if this map has a name, but let $q:S\to\mathcal{P}(F)$ be defined by $q(x)=\{A\in F:x\in A\}$. This induces a preorder on $S$ via $x\le y\iff q(x)\subseteq q(y)$. Preorder means $x\le x$ and $x\le y\land y\le z\rightarrow x\le z$. Like any preorder this induces an equivalence relation on $S$ via $x\sim y\iff q(x)=q(y)$. This equivalence relation is used in combinatorics of infinite sets. For instance, in this paper it is used in Theorem 3 when $F$ is indexed by an ordinal. After some polishing the map can be used to prove Kuratowski's theorem that any injection $\mathbb{N}\to\mathcal{P}(S)$ yields a surjection $S\to \mathbb{N}$ as explained here. I don't know what you need this for, but the induced equivalence relation is not a stranger.

Answer 2

The relation $\Rightarrow_F$ which you claim is a partial order, isn't a partial order. Here's a counterexample.

Let $S$ be $\{0, 1, 2\}$ and let $F$ be $\{\{2\}\}$. Now let $x$ and $y$ be $0$ and $1$ respectively. The following is true:

$$\{A\in F:x\in A\} = \{A\in \{\{2\}\}:0\in A\} = \varnothing$$ $$\{A\in F:y\in A\} = \{A\in \{\{2\}\}:1\in A\} = \varnothing$$

Therefore $x\Rightarrow_F y$ and $y\Rightarrow_F x$, but $x$ and $y$ are not the same.