Converse to uniqueness of continuous functions defined on dense sets Let $X$ and $Y$ be topological spaces with $Y$ Hausdorff. We know that if $A \subseteq X$ is dense, then whenever any two continuous functions $f, g: X \to Y$ agree on $A$, they must be identically equal.
I am wondering if the converse to this is true? That is, if there is a set $A \subseteq X$ with the property that whenever any two continuous $f$ and $g$ agree on $A$ implies they are equal, then we can conclude that $A$ is dense in $X$?

If $X$ is Tychonoff (completely regular) than this will hold, even with $Y$ fixed to $\Bbb R$ (instead of an arbitrary Hausdorff $Y$): suppose $A \subseteq X$ is not dense, then this means that we have some $p \in X\setminus \overline{A}$ and then we have (by complete regularity) a continuous $f: X \to \Bbb R$ such that $f(x)=0$ for $x \in \overline{A}$ and $f(p)=1$. Then $f$ and the constant real function with value $0$ agree on $A$ but not on $X$, so that $A$ fails the criterion.
So if $X$ is completely regular then any subset $A$ that obeys your property is indeed dense in $X$. Which is a relatively mild condition on $X$ (almost all spaces occurring in practice are completely regular).
In general we could define, for some fixed Hausdorff space $Y$, a set $A \subseteq X$ to be "functionally $Y$-dense" iff

for all pairs of continuous $f,g: X \to Y$, if $f\restriction_A = g\restriction_A$, then $f=g$.

And your first result is that a dense set $A$ is functionally $Y$-dense for any Hausdorff $Y$ and the converse can then be asked as follows: If $A$ is (non-empty and) functionally $Y$-dense for all Hausdorff $Y$, is then $A$ dense in $X$?
I just showed for completely regular spaces $X$ this is indeed true, and we can even fix $Y$ to $\Bbb R$ (or $[0,1]$  etc.) But if we fix $Y$ to a one-point space (voidly Hausdorff) then any subset of $X$ is functionally $Y$-dense and the converse then only holds for indiscrete spaces (as there any non-empty set is dense too). But in the cofinite topology on $\Bbb N$, all non-empty subsets $A$ are functionally $Y$-dense for any Hausdorff $Y$ (as all continuous functions into $Y$ are constant!) and a finite subset of $X$ is then a counterexample to the converse. Plenty to explore for other simple classes of spaces...
