Proving independence of the neighborhood axioms in topology

Question

I have not found any information on this topic on StackExchange or through a few minutes of searching with Google.

This question is Exercise 8 in Section 2.1 on page 22 of Topology and Groupoids, by Brown. I am given the neighborhood axioms

1. If $N$ is a neighborhood of $x$, then $x \in N$.
2. If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$.
3. The intersection of two neighbourhoods of $x$ is again a neighbourhood of $x$.
4. Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$.

and I am asked to prove that these four axioms are independent. It is my understanding that this can be accomplished by selecting selecting any subset of three axioms and supplying a purported neighborhood topology $\mathcal{N}$ that satisfies the all the axioms except the one that hasn't been selected. This requires $\binom{4}{3} = 4$ examples.

I think I have one example. If I say that $N$ is a neighborhood of $x \in X$ if and only if $N = \{ x \}$, then I think this satisfies axioms 1, 3 and 4 but not 2.

I haven't come up with any other examples yet, though. I looked briefly at Counterexamples in Topology by Steen and Seebach but it didn't look like they deal with this type of question.

Does anyone know of any other examples that will work? Thanks.

Edit:

I think I have another example. Let $X$ = $[0, 1]$ and say that $N$ is a neighborhood of $x \in X$ if and only if $\frac{1}{2} \in N$, then I think that satisfies axioms 2, 3 and 4 but not 1.

If I'm correct about both of these examples then I'm halfway to a solution.

Answer 1

For (1) you can let $X$ be any infinite set, and for each $x\in X$ let

$$\mathscr{N}(x)=\{X\setminus F:F\text{ is a finite subset of }X\}\;;$$

I’ll leave it to you to verify that (2)-(4) are satisfied and (1) is not.

For (3) let $X=\{0,1,2\}$. Let $\mathscr{N}(0)=\big\{\{0,1\},X\big\}$, $\mathscr{N}(1)=\big\{\{0,1\},\{1,2\},X\big\}$, and $\mathscr{N}(2)=\big\{\{1,2\},X\big\}$.

For (4) you can again let $X=\{0,1,2\}$, but this time let $\mathscr{N}(0)=\big\{\{0,1\},X\big\}$, and $\mathscr{N}(1)=\mathscr{N}(2)=\big\{\{1,2\},X\big\}$.