How to prove the equivalence of these two given definitions of neighborhood axioms？ Exercise 2.2.12 in Ronald Brown's "Topology and Groupoids":

The neighborhood axiom.

It's simple to proof N1~N3, but I don't know how to proof N4
Proof:
N1， from iii
$$\{x\}\lhd A\Rightarrow \subset A\Rightarrow x\in A\\$$
N2,
$$\{x\} \subset \{x\} \lhd A\subset B \Longrightarrow \{x\} \lhd B\\$$
N3,
$$\{x\} \lhd A, \{x\} \lhd B \Longrightarrow \{x\} =\{x\} \cap\{x\} \lhd A\cap B\\$$
But for N4, we need to proof
$$\forall A, \{x\} \lhd A, \exists B, \{x\} \lhd B \lhd A\\$$
I don't know how to do that, please help.
Is that equivalent? Am I misunderstand anything?
It seems wrong, but I am not sure, here is a example below.
Let $X$ be a infinite set，choose $x\in X$ and $x \in N \subset X, N \ne \{x\} \text{ and } N \ne X$, define $\lhd$ on some pairs,
$$A\lhd B := A=\varnothing \text{ or } B=X \text{ or } (A=\{x\} \text{ and } N \subset B)\\$$
We can see that this $\lhd$ satisfy the condition on execise 12, but for $x$  and its neighbors N, there isnt have any M for which N4 holds.
 A: Great, your counterexample to Ronnie Brown's exercise is correct!
It seems that he erroneously believed that the axioms define a topology such that $A \lhd B$ iff $A \subset \operatorname{Int} B$. If you think about it, it would be odd if it were true. As you say, to verify N4 one has to show that if $\{x\} \lhd A$, then there exist $M \subset A$ such that $\{x\} \lhd M$ and for all $x' \in M$ one has $\{x'\} \lhd N$. By axioms (ii) and (v) this is equivalent to finding $M$ such that $\{x\} \lhd M \lhd N$. But none of the axioms ensures that such a nesting is possible. We shall see that Ronnie Brown indeed forgot to add a "nesting axiom".
The purpose of the exercise it to assign to $\lhd$ a topology $\tau$ on $X$ such that
$$A \lhd B \Longleftrightarrow A \subset \operatorname{Int}_{\tau} B . \tag{1}$$
Let us call such $\tau$ a topological representation of $\lhd$. If $\tau$ has this property, then $A \in \tau \Leftrightarrow A \subset \operatorname{Int}_{\tau} A \Leftrightarrow A \lhd A$. The first equivalence is of course true for each topology. Therefore the only topology on $X$ that has a chance to be a topological representation of $\lhd$ is
$$\tau_\lhd = \{ A \subset X \mid A \lhd A \} . \tag{2}$$
This is indeed a topology on $X$; to show this we only need axioms (i), (iv) and (v).
$\tau_\lhd$ satisfies the implication
$$A \subset \operatorname{Int}_{\tau_\lhd} B \implies A \lhd B .$$
This follows from axiom (ii) because $\operatorname{Int}_{\tau_\lhd} B \in \tau_\lhd$, thus $A \subset \operatorname{Int}_{\tau_\lhd} B \lhd \operatorname{Int}_{\tau_\lhd} B \subset B$.
But is $\tau_\lhd$ a topological representation of $\lhd$?
To address this question, we consider a topology $\tau$ on $X$ and define
$$A \lhd_\tau B \Longleftrightarrow A \subset \operatorname{Int}_\tau B . \tag{3}$$
$\lhd_\tau$ satisfies all five axioms. To show this we only have to know that $\operatorname{Int}_\tau (B \cap B') = \operatorname{Int}_\tau B \cap \operatorname{Int}_\tau B'$ and $\bigcup_i \operatorname{Int}_\tau  B_i \subset  \operatorname{Int}_\tau \bigcup_i B_i$.
The assigment $\tau \mapsto \lhd_\tau \mapsto \tau_{\lhd_\tau}$ gives us the original topology because $A \in \tau_{\lhd_\tau} \Longleftrightarrow A \lhd_\tau A \Leftrightarrow A \subset \operatorname{Int}_\tau A \Leftrightarrow A \in \tau$.
What about the assigment $\lhd \mapsto \tau_\lhd \mapsto \lhd_{\tau_\lhd}$? By the above definitions we have

*

*$\tau_\lhd$ a topological representation of $\lhd$ if and only if $\lhd_{\tau_\lhd} = \lhd$.

Let us now define
$$B^\lhd = \{ x \in X \mid \{x\} \lhd B \} = \{ x \in B \mid \{x\} \lhd B \} . \tag{4}$$
Note that axiom (iii) shows that $\{x\} \lhd B$ is possible only for $x \in  B$. We have

*

*$A \lhd B \Longleftrightarrow A \subset B^\lhd$. Thus in particular $B^\lhd \lhd B$.

By axioms (ii) and (v) we have $A \lhd B$ iff $\{x\} \lhd B$ for all $x \in A$. The latter is equivalent to $x \in B^\lhd$ for all $x \in A$, i.e. to $A \subset B^\lhd$.
Since $\operatorname{Int}_{\tau_\lhd} B \in \tau_\lhd$, we have $\operatorname{Int}_{\tau_\lhd} B \lhd \operatorname{Int}_{\tau_\lhd} B$ and since $\operatorname{Int}_{\tau_\lhd} B \subset B$, axiom (ii) implies $\operatorname{Int}_{\tau_\lhd} B \lhd B$. Therefore

*

*$\operatorname{Int}_{\tau_\lhd} B \subset \ B^\lhd$.

But do we have $\operatorname{Int}_{\tau_\lhd} B = \ B^\lhd$ for all $B$?
Theorem. The following are equivalent:

*

*$\tau_\lhd$ is a topological representation of $\lhd$.


*$\operatorname{Int}_{\tau_\lhd} B = \ B^\lhd$ for all $B$.


*$B^\lhd \in \tau_\lhd$ (i.e. $B^\lhd \lhd B^\lhd$) for all $B$.


*$\lhd$ satisfies the following nesting axiom:
(vi) For all $A, B$ such that $A \lhd B$ there exists $C$ such that $A \lhd C \lhd B$.
Proof. 1. $\implies$ 2. :
We have to show that $B^\lhd \subset \operatorname{Int}_{\tau_\lhd} B$. Since $\tau_\lhd$ is a topological representation of $\lhd$, this is equivalent to $B^\lhd  \lhd B$. But it was shown above that this is always satisfied.


*$\implies$ 3. is trivial.


*$\implies$ 4. :
We know that if $A \lhd B$, then $A \subset B^\lhd$. Since $A \subset  B^\lhd \lhd B^\lhd \subset B$, axiom (ii) shows that $A \lhd B^\lhd \lhd B$.


*$\implies$ 1. :

It remains to prove the implication $A \lhd B \implies A \subset \operatorname{Int}_{\tau_\lhd} B$. Let $\mathcal C$ denote the set of all $C$ with $A \lhd C \lhd B$. Define $C^*  = \bigcup_{C \in  \mathcal C} C$. Clearly $A \subset C^* \subset B$. We claim that $C^* \lhd C^*$; this means that $C^* \in \tau_\lhd$, thus $C^* \subset \operatorname{Int}_{\tau_\lhd} B$. By axiom (v) it suffices to show that $C \lhd C^*$ for all $C \in \mathcal C$. But since $C \lhd B$, we find $C'$ such that $C \lhd C' \lhd
 B$. Thus $C' \in \mathcal C$ and $C \lhd C' \subset C^*$ so that axiom (ii) applies.
Let us observe that $\lhd_\tau$ from definition $(3)$ satisfies axiom (vi): Take $C = \operatorname{Int}_{\tau} B$. Then $C  = \operatorname{Int}_{\tau} C = \operatorname{Int}_{\tau} B$ which shows that $A \subset \operatorname{Int}_{\tau} C$ and $C \subset \operatorname{Int}_{\tau} B$.
Axiom (vi) is not satisfied in your counterexample, thus it is independent from axioms (i) - (v). One can easily construct more counterexamples violating axiom (vi).
In the above considerations we did not use Ronnie Brown's suggested approach by associating to $\lhd$ a neighborhood topology $\mathcal N$. But clearly axiom (vi) implies N4.
It is well-known that the concepts of topology and neighborhood topology are equivalent.
