The following is Exercise 2.2.12 is Ronald Brown’s Topology and Groupoids, which tries to characterize topological spaces using the relation “$A$ is contained in the interior of $B$”.
Let $X$ be a non-empty set and $⊲$ a relation on subsets of $X$ such that
$∅ ⊲ ∅$ and $X ⊲ X$
$A ⊆ A'$, $A' ⊲ B'$, $B' ⊆ B$ imply $A ⊲ B$.
$A ⊲ B$ implies $A ⊆ B$.
$A ⊲ B$ and $A' ⊲ B'$ imply $A ∩ A' ⊲ B ∩ B'$.
$A_i ⊲ B_i$ for all $i ∈ I$ implies $⋃_{i ∈ I} A_i ⊲ ⋃_{i ∈ I} B_i$.
For each $x ∈ X$, $A ⊆ X$ define $A$ to be a neighbourhood of $x$ if $\{ x \} ⊲ A$. Prove that these neighbourhoods define a topology on $X$ for which $A ⊲ B$ if and only if $A ⊆ \operatorname{Int}(B)$.
By a topology the book means a neighbourhood topology in the following sense:
Let $X$ be a set and let $\mathcal{N}$ be a function assigning to each $x$ in $X$ a set $\mathcal{N}(x)$ of subsets of $X$. The elements of $\mathcal{N}(x)$ will be called neighbourhoods of $x$. The function $\mathcal{N}$ is a neighbourhood topology if it satisfies the following axioms for each $x$ in $X$:
The set $\mathcal{N}(x)$ is non-empty.
If $N$ is a neighbourhood of $x$, then $x ∈ N$.
If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$.
The intersection of any two neighbourhoods of $x$ is again a neighbourhood of $x$.
Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.
My attempts so far:
It is not hard to see that properties 0–3 of a neighbourhood topology are satisfied, but I’m struggling to show property 4. Given a neighbourhood $N$ of $x$ we can consider the set $$ M = \{ y ∈ N \mid \{ y \} ⊲ N \} $$ and this should a posteriori do the job. But I don’t see why $M$ should be a neighbourhood of $x$. I also tried to modify this approach by defining $N^{(0)} ≔ N$, $N^{(1)} ≔ M$, etc. and then considering $⋂_{n = 0}^∞ N^{(0)}$. But this doesn’t seem to work either.
I also tried to solve the exercise the other way around: It follows from the properties of the relation $⊲$ that the set $$ \mathcal{U} ≔ \{ A ⊆ X \mid A ⊲ A \} $$ forms a topology in the usual sense on $X$ (i.e., a collection of “open subsets” satisfying certain conditions). Is is then not hard to see that $A ⊆ \operatorname{Int}(B)$ implies $A ⊲ B$, but I don’t see how the reverse implication is supposed to work.
I suspect by now that the exercise may be missing some additional requirement on the relation $⊲$. There seems to be the more general notion of a pretopological space for which axiom 4 of a neighbourhood topology is not required. In such a space $X$, one can still define the preinterior of a subset $A$ of $X$ as $$ \operatorname{PreInt}(A) = \{ x ∈ X \mid \text{$A$ is a neighbourhood of $x$} \}. $$ It seems to me that the relation $⊲$ given by $A ⊲ B$ iff $A ⊆ \operatorname{PreInt}(B)$ should satisfy all of the given conditions. But the resulting “topology” is just the original pretopology.
