Understanding axioms of neighbourhoods in topological spaces Let $X$ be a set of points, along with a certain function $N$ that assigns a subset to every $x \in X.$ Then $X$ together with $N$ define a topological space if $N$ satisfies the following axioms: (quoting below from wikipedia)

  
*
  
*If $N$ is a neighbourhood of $x$ (i.e., $N \in N(x)$), then $x \in N$. In other words, each point belongs to every one of its neighbourhoods.
  
*If $N$ is a subset of $X$ and contains a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. I.e., every superset of a neighbourhood of a
  point $x$ in $X$ is again a neighbourhood of $x$.
  
*The intersection of two neighbourhoods of $x$ is 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 $M$.
  

Questions:


*

*My first struggle is with the statement of the second axiom, namely that any superset $S\supseteq M$ of a neighbourhood $M$ of $x$ is again a neighbourhood of $x$. 


*

*a) Doesn't this mean that everything is a neighbourhood of everything because I can just set $S=X$? But then what good $N$ is at characterizing $X$ in terms of specific subsets of every $x$, if any neighbourhood can be indefinitely enlarged?

*b) How to understand the need for such an axiom?


*Am I right to interpret the 4th axiom as follows: Given a neighbourhood $N$ of $x,$ it is always possible to extract a smaller neighbourhood $M'\subseteq M$  from $M,$ such that $M'$ is an open subset, where $M$ need not be open itself.

*To me it seems that the 2nd and 4th axioms are equivalent statements. What makes them distinct? (or in other words what makes the 4th axiom necessary).
 A: The enlargement axiom does not trivialise the theory. It does imply that $X$ is a neighbourhood of every $x \in X$: every $x \in X$ has some $N \in N(x)$ (this is actually part of the axioms/conditions, often forgotten: every $N(x)$ is non-empty) and $N \subseteq X$ so $X \in N(x)$ by axiom 2. And everything inbetween $N$ and $X$ is a neighbourhood too. We cannot go smaller in general, so it's not true that everything becomes a neighbourhood of everything: $X$ is a neighbourhood of every point, but any fixed set $A$ is only a neighbourhood of $x$ if it happens to be the case that $A \in N(x)$, and otherwise it's not. 
As to the necessity of the axiom: look at metric spaces, where $A$ is a neighbourhood of $x$ iff there is some $r > 0$ such that $B(x,r) \subseteq A$. The way this is defined already implies that larger sets than $A$ are also neighbourhoods of $x$; the same $r$ will work. So it's not a very surprising axiom. The same can be said for ordered spaces ($A$ is a neighbourhood of $x$iff either there exists $a \in X$ with $x < a$ and $\{z : z < a\} \subseteq A$, or there exists $b \in X$ with $b < x$ and $\{z : b < z \} \subseteq A$ or there exist $a,b \in X$ with $a < x, x < b$ and $\{z: a < z, z < b \} \subseteq A$), where also the enlargement is clearly true. We really are interested in the "small" neighbourhoods of $x$ (for continuity etc.) but in this axiomatisation we want to capture the notion of all possible neighbourhoods, even the not so interesting large ones. We could also axiomatise "basic" neighbourhoods instead, instead of all neighbourhoods, but this is not what we are doing here. In short, the relation of $A$ being a neighbourhood of $x$ is "local" and not affected by enlarging the set.
As to axiom 4, it is true that the "intended" result is that every neighbourhood of $x$ contains an open neighbourhood of $x$, but for that we first need to define what "open" means. We only have a set of sets $\{N(x): x \in X\}$ satisfying the  axioms, no notion of openness. But if you already know the standard axioms for topology in terms of open sets, you want to go back and forth: 
If you have a topology $\mathcal{T}$, define for each $x \in X$: $N(x) = \{A \subseteq X: \exists O \in \mathcal{T}: x \in O \subseteq A\}$ and check that this fulfills the axioms (again see that 2 is trivially fulfilled). Call this system $\mathcal{N}(\mathcal{T})$. On the other hand, if we have a system $\mathcal{N}= \{N(x): x \in X \}$ satisfying the axioms, define $\mathcal{T}  = \{O \subseteq X: \forall x \in O: O \in N(x) \}$ and check that this defines a topology (we do not need axiom 4 for this, it turns out). Call this topology $\mathcal{T}(\mathcal{N})$. The axioms 4 is used to show that $\mathcal{N}(\mathcal{T}(\mathcal{N})) = \mathcal{N}$ again. So the neighbourhood system from the topology induced by the neighbourhood system is that same one we started with. It's a fun exercise to check this out in detail. It will enhance your understanding of these axioms. It's also true that $\mathcal{T}(\mathcal{N}(\mathcal{T})) = \mathcal{T}$, which is also fun to prove.
