Let $(X, \tau)$ be a topological space and $\tau$ be the set of opens.
Let's call a pair of the form $(x, W)$ where $x$ is a point in $X$ and $W$ is a (not necessarily open) neighborhood of $x$ a neighborhood pair.
Let's call the collection of all such neighborhood pairs $N$.
It seems pretty clear that we can convert back and forth between $(X, \tau)$ and $(X, N)$ without losing any information. An open set $U$ is a set that is a neighborhood of each of its points. The same token, a set $T$ is a neighborhood of $x$ if and only if it contains an open set $U$ that contains $x$.
I'm wondering what happens when you strip away the information about what point a neighborhood is a neighborhood of, i.e.
$$ N \rightsquigarrow \{ W : (x, W) \in N \} $$
Call $N$ the set of neighborhoods corresponding to the topological space in question.
Can we recover our topology from just the set of neighborhoods, neglecting completely which points they are neighborhoods of?.
We definitely have a way of getting an answer by just intersecting a ton of topologies.
Given a family of topologies $F$ over a shared point set $X$, $\cap F$ is a topology.
For proof, see here.
Alternatively, let's check the axioms one at a time.
- $\varnothing$ is in $\cap F$.
- $X$ is in $\cap F$.
- if $A$ is in $\cap F$ and $B$ is in $\cap F$, then $A$ and $B$ are in each $\tau \in F$ individually, thus $A \cap B$ is in each $\tau$ individually, thus $A \cap B$ is in $\cap F$.
- Suppose $E \subset \cap F$, then $E \subset \tau$ for each $\tau$ in $F$. Thus $\cup E$ is in each $\tau$, thus $\cup E$ is in $\cap F$.
Suppose our set of neighborhoods is $M$.
Let $\Lambda$ be the family of all topologies on $X$.
Let $L \subset \Lambda$ be the family of all topologies on $X$ whose set of neighborhoods is a superset of $M$. $L$ is not empty. The discrete topology on $X$ will always be in $L$.
Let's call $\cap L$ $\sigma$. $\sigma$ is a topology.
There's no guarantee though that the set of neighborhoods of $(X, \sigma)$ will be $M$ though and no guarantee that $(X, \sigma)$ will be the topology we started with to produce $M$.
- This construction always succeeds, regardless of whether it's possible to get a topology whose set of neighborhoods is $M$ or not.
- If there are multiple topologies whose set of neighborhoods is $M$, this construction will give us the coarsest one.
Best Answer
No. Your set $N$, the neighborhoods, are just all possible supersets of non-empty open sets and this collection may coincide for different topologies. For example, let $X=\{1, 2, 3\}$ and consider the following topologies:
$$\tau_1=\{\emptyset, \{1\}, X\},$$ $$\tau_2=\{\emptyset, \{1\}, \{1, 2\}, X\},$$ $$\tau_3=\{\emptyset, \{1\}, \{1, 2\}, \{1, 3\}, X\}.$$
Then for any one of the topologies the neighborhoods are just the sets that contain $1$. Another example would be the real line with the standard topology and the real line with the lower limit topology, i.e. where a basis of topology are the intervals $[a,b)$, $a<b\in \mathbb{R}$. Indeed, one has $$[a,b)\supseteq (a, b)$$ and $$(a, b)\supseteq \left[a+ \frac{b-a}{2}, b\right)$$ for all $a<b\in \mathbb{R}$, showing that the supersets of the respective bases, which are the respective neighborhoods, coincide.