Why Does the Definition of a Topology via Neighborhoods Include This Axiom

Question

My question concerns the Definition via Neighborhoods of a topological space. I am having trouble seeing why $(4)$ is necessary in the definition, which I'll briefly run through below.

---

Let $X$ be a set and define a function $\mathcal{N}:X \to \mathcal{P}(X)$ which assigns to each $x \in X$ a nonempty collection of subsets of $X$, which we call neighborhoods of $x$, satisfying the following properties:

1. If $N \in \mathcal{N}(x)$ then $x \in N$.
2. If $N,M \in \mathcal{N}(x)$ then $N\cap M \in \mathcal{N}(x)$.
3. If $N \in \mathcal{N}(x)$ then any set $Y\supseteq N$ is also in $\mathcal{N}(x)$.
4. If $N \in \mathcal{N}(x)$ then there exists an $M \subseteq N$ in $\mathcal{N}(x)$ such that $N \in \mathcal{N}(y)$ for all $y \in M$.

Define a set $U \subseteq X$ to be open in $X$ if $U$ is a neighborhood of all of its points.

---

As I mentioned above, I am not sure why we need $(4)$ because (I think) I can recover the standard definition of a topology via open sets just from $(1)$ through $(3)$. Assuming the definition of an open set given above we see that clearly $\varnothing$ is open because it is vacuously a neighborhood of all of its points. Moreover, $X$ is open by $(3)$. Finite intersections of open sets are open by $(2)$, and arbitrary unions of open sets are open by $(3)$.

Wikipedia mentions that $(4)$ "has a very important use in the structure of the theory", but I don't see how this is possible if I can recover the standard definition of a topology via open sets without it.

I feel that I must be missing something simple. Have I made a mistake? Do you actually need $(4)$ to recover the standard definition of a topology via open sets? If not, then how is $(4)$ important?

Answer 1

1. When you define topological terms using open sets as a foundation, open sets are defined axiomatically and neighborhoods are defined as any set containing an open set.

2. When you define topological terms using neighborhoods as a foundation, neighborhoods are defined axiomatically and open sets are defined as any set which is a neighborhood of all of its points.

3. These definitions coincide if you use the four axioms above. As you point out, the first three are enough to make sure that open set means the same thing in both scenarios. The fourth axiom is necessary to make sure that neighborhood means the same thing.

4. If you exclude the fourth axiom, then your definition might allow a neighborhood (axiomatically defined) which is not a neighborhood (a set containing an open set).

   For example, take the real line $\mathbb{R}$ and define a neighborhood system axiomatically as follows. For each point $x\neq 0$, let $\mathbb{R}$ be its only neighborhood. Let the interval $(-1,1)$ and its supersets be the neighborhoods of $x=0$. You can show that the only open sets are $\mathbb{R}$ and $\varnothing$ ^{[*proof below]}. It follows that $(-1, 1)$ is a neighborhood of 0 according to neighborhood axioms, but not a neighborhood of 0 according to open set axioms— it contains only the trivial open set $\varnothing$.

   ^[*] Proof of claim: if $U$ is an open set and $U$ contains $x\neq 0$, then $U=\mathbb{R}%$ because $\mathbb{R}$ is the only neighborhood of $x$. If $U$ contains only 0, then $U=\{0\}$, which isn't an open set because it isn't a neighborhood. If $U$ is empty, then $U$ is trivially an open set.

Answer 2

You need it because the intended meaning for neighbourhood of $x$ is that it contains an open set containing $x$. Also, we want $O$ open to mean "$O$ is a neighbourhood of each of its points". Axiom (4) ties this together, and also guarantees that when we define a topology from the neighbourhood system, that the neighbourhoods for that newly defined topology are exactly the neighbourhoods we started out with; it gives a bijection between neighbourhood systems and topologies.