Understanding the axioms for neighbourhoods and their independence/consistency

Question

After reading answers to this question I can't help being confused about axioms 2 and 4 being truly independent, or about their true meaning for that matter.

Recalling them, for a given set $X$:

1. If $N$ is a neighbourhood of $x \in X$ (i.e., $N \in \mathcal N(x)$), then $x \in N$.
2. If $N \in \mathcal N(x)$ and $N \subseteq N'$ then $N' \in \mathcal N(x)$.
3. If $N, N' \in \mathcal N(x)$ then $N \cap N' \in \mathcal N(x)$.
4. For any $x \in X$ and $N \in \mathcal N(x)$ there exists $N' \in \mathcal N(x)$ such that for every $y \in N'$ it holds that $N \in \mathcal N(y)$.

Now my questions:

1. By (1) and (4), does it follow that $N' \subseteq N$ (referring to $N'$ and $N$ as in 4)?
2. Are $N'$ and $N$ required to be different (i.e. 'proper inclusion') in 4?
3. Can you give an example of a neighbourhood system that would satisfy the first three axioms but not the fourth (assuming the later false for the purpose)?

Answer 1

I find that it helps with understanding these types of statements to firstly re-state them in more plain English. So "if $N$ is a neighbourhood of $x$, then if $N$ is enlarged it remains a neighbourhood of $x$" and "if $N$ is a neighbourhood of $x$, then it contains a smaller neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point in $M$". It also helps to have some (perhaps slightly non-rigorous) heuristic, or slogan, or picture in mind. For example "$N$ is a neighbourhood of $x$ means that if you move $x$ a tiny little bit then it stays in $N$" (then axiom 4 says "if you move $x$ a tiny bit, $N$ will still be a neighbourhood of $x$"). Or you might find it helps you to ground these examples in a concrete setting like $X = \Bbb R^2$ with the usual topology, by checking why these axioms are actually true in that case.

As for your questions:

1. Yes, indeed the $N'$ in axiom 4 will be a subset of $N$ (in fact that's even how it's stated at the question you link).

2. No, that's not required. For example, in a space with the indiscrete topology, every point has only a single neighbourhood.

3. To give an example of axiom 4 failing, we have to give an example of a set and neighbourhood systems for all points in the set. Axiom 4 is the only axiom that says anything about how neighbourhood systems of different points interact with each other - so if you "randomly" assign neighbourhood systems which are otherwise plausible, you're likely to come up with a counterexample.

   Here is one: let $X$ be $\Bbb R$. Let $\mathcal N(0)$ be the set of all neighbourhoods in the usual sense (ie all subsets of $\Bbb R$ that contain $(-\epsilon, \epsilon)$ for some $\epsilon > 0$). For any $x \ne 0$, let $\mathcal N(x)$ consist only of the set $\Bbb R$. Roughly, we're mashing together the notions of "neighbourhood" from the Euclidean topology and from the indiscrete topology, and hoping something will go wrong. For each $x$, axioms 1-3 are satisfied. However, for example, there is no neighbourhood $M$ of $0$ contained in $(-1, 1)$ such that $(-1, 1)$ is a neighbourhood of every point in $M$. So axiom 4 is violated.

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One can come up with more extreme examples, where axiom 4 is violated "everywhere". The first one that I can come up with off the top of my head is this: take $X = \Bbb R$ again. For each $x \in \Bbb R$, let $\mathcal N(x)$ be the set of all subsets of $\Bbb R$ that contain $(x - 1, x + 1)$. It's again not hard to see that axioms 1-3 are satisfied everywhere. However, for each $x$, the neighbourhood $(x - 1, x + 1)$ is only a neighbourhood of $x$, but its only subset which is a neighbourhood of $x$ is $(x - 1, x + 1)$. So axiom 4 now fails "everywhere".

There is no real need to use $\Bbb R$, of course. This example can be made to work just as well using a similar idea with a finite set.