Continuity of a function through adherence of subsets We say two sets $A,B$ being $adherents$ if we have $(\overline{A} \cap B)\cup(A\cap\overline{B})\neq \emptyset $.
Prove that  a function $f:X\to Y$, with $X,Y$ topological spaces, is continuous if and only if for  every $A,B$ adherents in $X$ then $f(A),f(B)$ are adherents in $Y$.
 A: I think I have a counterexample. Let 
$$X=(\{a,b\},\{\emptyset,\{a,b\},\{a\}\}),\quad 
Y=(\{a,b\},\{\emptyset,\{a,b\},\{b\}\}),\quad f=Id:X\to Y$$
Then $\overline{\{a\}}\cap\{b\}=\{b\}$ and $f(\{a\})\cap\overline{f(\{b\})}=\{a\}$, so adherents map to adherents, but $f$ is not continuous.
A: To extend the answer by Stefan Hamcke, which gives probably the best and minimal counterexample:
Function $f$ is continuous iff $x ∈ \overline{A} \implies f(x) ∈ \overline{f(A)}$, equivalently $\overline{A} ∩ B ≠ ∅ \implies \overline{f(A)} ∩ f(B) ≠ ∅$, which is asymetric (and stronger) variant of your condition. So continuity implies your condition.
We call a topological $X$ space symmetric or $R_0$ iff $x ∈ \overline{\{y\}} \iff y ∈ \overline{\{x\}}$ for $x, y ∈ X$. Note that a topological space is $T_1$ iff it is symmetric and $T_0$.
Your condition implies continuity if the target space is symmetric. Let $x ∈ \overline{A}$, then by the condition $\{f(x)\}$ and $f(A)$ are adherent in $Y$. So either $f(x) ∈ \overline{f(A)}$, which we want, or there is some $y ∈ f(A) ∩ \overline{\{f(x)\}}$. By symmetry $f(x) ∈ \overline{\{y\}} ⊆ \overline{f(A)}$ and we are done.
