Here is an ad hoc attempt. A connectivity space is an ordered pair $\ \mathbf X:=(X\ \mathcal C)\ $ such that the following two axioms hold: 1. $\ \left(A\ne\emptyset\ne B\,\ \wedge\ \,A\ B\in\mathcal C\right)\\ \quad \Rightarrow\quad \left( A\cup B\in \mathcal C\ \ \Leftrightarrow\quad\exists_{x\in A\cup B} \left(A\cup\{x\}\in\mathcal C\ \ \wedge\ \ B\cup\{x\}\in \mathcal C\right)\ \right) $ 2. $\ \forall_{x\ y\,\in\,A}\ \exists_{S\in\mathcal C}\ (x\ y\in S\ \ \wedge\ \ S\subseteq A)\quad\Rightarrow\quad A\in \mathcal C$ for every $\ A\ B\ \subseteq X.\ $ Next, given connectivity spaces $\ \mathbf X:=(X\ \mathcal C)\ $ and $\ \mathbf Y:=(Y\ \mathcal D),\ $ A connectivity map (or connectivity morphism) is any function $\ f:X\rightarrow Y\ $ such that $\ \forall_{A\in\mathcal C}\ f(A)\in\mathcal D.$ Given a topological space $\ \mathbf X:=(X\ T),\ $ we get the induced connectivity space $\ \mathbf X_c := (X\ \mathcal C_T),\ $ where $\ C_T\ $ is the family of all connected subsets of $\ \mathbf X$. Thus every continuous map between two topological spaces is a connectivity map between the induced connectivity spaces. Do not expect that there is a very close relation between continuous maps and connectivity maps. But the relation between them should be interesting (a source of new MO-questions :-) ). > **REMARK** It follows from the above definition (two axioms) that $\ \emptyset\in\mathcal C.\ $ (*Thank you Eric for this point*). **EXAMPLE** An intersection $\ \bigcap_{n=1}^\infty A_n\ $of a monotone sequence of closed connected subspaces $\ A_n\ $ doesn't have to be connected. For instance, consider the following subspaces of $\ \mathbb R^2$: $$A_n\ :=\ \mathbb R^2\setminus (-1;1)\times(-n;n)$$ for $\ n=1\ 2\ \ldots$ Of course a small modification will give a similar example for open connected subsets $\ B_n,\ $ say: $$B_n\ :=\ \mathbb R^2\setminus \{0\}\times[-n;n]$$ >A point-free definition: A connectivity structure $\ \mathcal C\ $ in set $\ X,\ $ with $\ \mathcal C_0:=\mathcal C\setminus \{\emptyset\},\ $ is defined by the following 3 axioms: - $\forall_{A\ B\,\in\,\mathcal C_0}\ \left(A\cup B\in\mathcal C_0\quad\Leftrightarrow\quad\exists_{S\in\mathcal C_0}\left(S\subseteq A\cup B\ \ \wedge\ \ A\cup S\in \mathcal C_0\ \ \wedge\ \ B\cup S\in\mathcal C_0\right)\ \right)$ - $\forall_{R\ S\,\in\,\mathcal C_0}\, \left(\left(R\cup S\,\subseteq\, A\right)\ \Rightarrow\ \exists_{Q\in\mathcal C_0}\ R\cup S\subseteq Q\subseteq A\ \right)\quad\Rightarrow\quad A\in\mathcal C_0$ - $\emptyset\,\in\,\mathcal C$ for every $\ A\ B\,\in\,X$.