Finite membership and its scope

For a complete catalogue, a finite separator excludes a global constant-coefficient identity. A finite fit has no converse global implication. For example, on $\mathbb{R}$, for any prescribed finite $S^{\prime }=\{t_1,\dots ,t_s\}$,

agree on $S^{\prime }$ but not globally.

One sufficient finite-to-global route is injective evaluation on

If evaluation on a family is injective on $U$, the difference of a fitted identity has zero evaluation and hence is the zero function. Such injectivity is not proved for $S$. A direct global proof would be another route.