Consider a set of points $\mathcal{A}$. I want to define the subset $\mathcal{A}^*$ of $\mathcal{A}$ as the collection of elements of $\mathcal{A}$ such that, for every two pairs of such elements $a,a'$, it holds that $$ f(a,a')=k \text{ implies } a=a' $$
This is a circular definition, as observed in the comment below.
1) Is such a circular definition admissible?
2) If it is admissible, how do we define formally $\mathcal{A}^*$?
My attempt is $$ \mathcal{A}^*\equiv \{a\in \mathcal{A}: \text{ }\forall a'\in \mathcal{A}^*,f(a,a')=k \text{ implies } a=a'\} $$
An arbitrary subset $S$ of $\mathcal{A}$ either does or doesn't have the property you seek for $\mathcal{A}^\ast$, namely that any $a,\,b\in S$ with $f(a,\,b)=k$ implies $a=b$. This property of $S$ defines a unique choice of $\mathcal{A}^\ast$ if, and only if, exactly one subset of $A$ has the desired property. This may be true or false depending on the choice of $\mathcal{A},\,f,\,k$. However, the empty set certainly does, so $\mathcal{A}$ is either undefined or the empty set.