Alternative to dense subsets for non-Hausdorff spaces

Question

Dense subsets of a topological space $X$ satisfying strong enough separation axioms (Hausdorff is enough) have the property that

for any two continuous maps $f:X \to Y$ and $g:X \to Y$ into any topological space $Y$, if $f = g$ on a dense subset of $X$, then $f=g$.

Is there a strengthening of the condition "dense", as a property of subsets of $X$, or maybe of "dense image" as a property of continuous maps, to make this true in arbitrary topological spaces? Certainly for any $X$, being a subset that uniquely determines the values of continuous functions, singles out a unique collection of subsets, so the question is whether there is a description of this property that is uniform with respect to $X$.

Answer 1

It’s $Y$ that has to be Hausdorff in order for that result to go through. If you don’t require $Y$ to be Hausdorff, then for a $T_1$ space $X$ the only subset that completely determines all continuous functions with domain $X$ is $X$ itself.

Let $\langle X,\tau\rangle$ be any $T_1$ space, and let $D$ be any proper subset of $X$. Fix $p\in X\setminus D$, let $q$ be any point not in $X$, let $Y=X\cup\{q\}$, and let

$$\tau'=\big\{Y,X,Y\setminus\{p\}\big\}\cup\{U\in\tau:p\notin U\}\;.$$

Let $f:X\to Y:x\mapsto x$ and

$$g:X\to Y:x\mapsto\begin{cases} q,&\text{if }x=p\\ x,&\text{if }x\ne p\;. \end{cases}$$

Then $f$ and $g$ are continuous, and $f\upharpoonright D=g\upharpoonright D$, but $f\ne g$.

Essentially I just changed the topology at $p$ to make $X$ its only nbhd and then added a second copy of $p$ to form $Y$.

More trivially, but also more generally, let $X$ be completely arbitrary, and let $Y$ be the two-point space with the indiscrete topology. Then every map from $X$ to $Y$ is continuous, so no proper subset of $X$ determines all continuous functions with domain $X$: you can always change the value at any one point without losing continuity.

Answer 2

First of all, for the proposition you stated to hold, you need that $Y$ is hausdorff, not $X$.

Next, I think what you're searching for is not possible. I'll make my point:

Take a topological space $X$, and denote by $Y$ any set with the trivial topology. Hence, every function $h: X \rightarrow Y$ is continuous. Now, it doesn't matter the subset $A$ we take, we can always get continuous functions (any function) satisfying the fact that $f |_A=g|_A$ but $f \neq g$ (if $Y$ has more than one element). Of course, we are assuming $A$ not to be all of $X$, or this would be trivial.

Making my point clearer: your "generalization" of dense (meaning: satisfying "unique potential-extendability") should be a topological condition on $X$, and should not depend on the $Y$. But by what I said above, no subset could satisfy "unique potential-extendability" of continuous functions.

With respect to your latter question:

or maybe of "dense image" as a property of continuous maps...

this is not clear: on what would you impose such property? For your first question, you are explicitly asking the condition to be imposed on a given $A \subset X$. Could you elaborate?