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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 $M$.

  1. This construction always succeeds, regardless of whether it's possible to get a topology whose set of neighborhoods is $M$ or not.
  2. If there are multiple topologies whose set of neighborhoods is $M$, this construction will give us the coarsest one.
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  • $\begingroup$ This is not quite true @spaceisdarkgreen, for example in my answer below, the set $\{1, 2\}$ meets that requirement in the topology $\tau_1$, even though it is not open in $\tau_1$. The correct statement is that $O$ is open iff for every $x\in O$ there is neighborhood of $x$ such that $x\in W\subseteq O$, but that doesn't help in recovering a topology just from the set of neighborhoods. $\endgroup$ Jan 1 at 8:05

2 Answers 2

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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.

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  • $\begingroup$ For your first example, how does the collection of neighbourhoods coincide? For, $\tau_1$ doesn't have the neighborhood $\{1,2\}$ where $\tau_2$ has. The 2nd example is alright. $\endgroup$
    – MAS
    Jan 2 at 18:21
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    $\begingroup$ @MAS: I struggle to understand how $\{1, 2\}$ avoids being a neighborhood of 1 in $\tau_1$, since it contains the open set $\{1\}$. But maybe I'm missing something obvious. $\endgroup$
    – Kevin
    Jan 2 at 18:48
  • $\begingroup$ @Kevin, yes. Because by definition, a neighborhood $N$ of a point $p$ in a topological space $(X,\tau)$ is a set such that there is an open set $U$ of $X$ for which $p\in U \subset N$. Since $\{1,2\}$ is an open, it is a neighborhood of 1 by taking $U=N=\{1,2\}$. I don't see where I am wrong. $\endgroup$
    – MAS
    Jan 3 at 0:14
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    $\begingroup$ @MAS In both topologies, $N = \{1,2\}$ is a neighborhood of $p = 1$, as it contains the open set $U = \{1\}$ which contains $1$. Therefore, $\tau_1$ and $\tau_2$ both have neighborhoods $\{\emptyset, \{1\}, \{1,2\}, X\}$. $\endgroup$
    – Cecilia
    Jan 3 at 0:36
  • $\begingroup$ @Cecilia, thanks I see it now. $\endgroup$
    – MAS
    Jan 3 at 1:10
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A "neighborhood" is a set whose interior is nonempty. I guess you are asking if we can determine the topology on a set if we know which subsets have nonempty interiors, which is the same as knowing which subsets are dense: a set has nonempty interior iff its complement is not dense.

Consider two topologies on the set $X=[0,2\pi)$; the standard topology as a subspace of the real line, and the topology which makes the map $t\mapsto e^{it}$ a homeomorphism from $X$ to the unit circle. Plainly, these are two different topologies with the same dense sets and therefore the same "neighborhoods". A more exotic topology on $X$, still with the same dense sets, is the topology generated by the half-open intervals of the form $[a,b)$.

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  • $\begingroup$ I edited my answer before I saw yours to include essentially the same example, so now it looks a bit weird. Oh well. $\endgroup$ Jan 1 at 10:41

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