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I have been thinking about two basic "equivalent" ways of defining a topological space and wanted to know if my reasoning is sound.

The first way of defining an abstract topological space is by assigning to points of a set $X$, collections of subsets of $X$ called neighborhoods satisfying

(1) Each $x\in X$ lies in each of its neighborhoods
(2) Any subset of $X$ containing a neighborhood of $x$ is also a neighborhood.
(3) The intersection of any two neighborhoods is a neighborhood.
(4) For any neighborhood of $x$, the set of all points that it is a neighborhood of is also a neighborhood

We call these neighborhoods and this assignment of neighborhoods to each point is called a topological space. Now, define open sets of the topological to be those that are neighborhoods of each of their points, where neighborhoods are subsets satisfying the above axioms. One can show that any union and any finite intersection of open sets as defined, is open, as neighborhoods satisfying the above axioms. To this end, the whole space $X$ is a neighborhood of all its points since it contains a neighborhood, namely any open set (this is (2)).

The other way one can define a topology is by saying that a topology on a set $X$ is an assignments of sets we call open, "closed" under taking unions and finite intersections. Then, define neighborhoods in terms of open sets: a neighborhood of a point is a subset $N\subseteq X$ for which there exists an open set $O$ such that $x\in O\subseteq X$. It can be shown that defining neighborhoods in this way gives a each point a collection of subsets satisfying the four axioms above. This is essentially where the equivalence of definitions comes from.

On one hand we call sets with closure of unions and finite intersections open sets and call that a topology on $X$ in the most axiomatic and abstract way. On the other hand, we start with primitive sets satisfying the 4 axioms, inspired by Euclidean metric spaces as a topology, then defining open sets in terms of these sets. Finally, starting with some open sets and using them to build a topology, the resulting open sets in the topology will match the original ones. Similarly, starting with a topological space based on neighborhoods satisfying the four axioms, then defining open sets and neighborhoods of each point as sets $N$ such that $O\subseteq N$ for some open set, the two notions of neighborhoods coincide.

I would like to know if my reasoning here is sound. If this is the case, my question is regarding whether or not, at the end of the day, the most concise way of introducing the notion of a topology is by introducing open sets as those sets closed under unions and finite intersections. Then neighborhoods as an artifact.

Thanks.

Thomas Andrews
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    What does "the set of all points that is a neighborhood is a neighborhood" mean? I think you mean, for a neighborhood $U$ of $x,$ the set $V$ of all $y$ such that $U$ is a neighborhood also of $y,$ is a neighborhood of $x.$ It took several reading because you avoid variables and it sounded like you were saying the individual points were neighborhoods - that is, that you were still describing the conditions,on the points of $V.$ That is the risk of reaching for too much brevity and avoiding notation. – Thomas Andrews Jan 27 '25 at 07:43
  • @ThomasAndrews "the set of all points that it is a neighborhood of is a neighborhood". I don't know if it's correct English for "the set of all points of which it is a neighbourhood is a neighbourhood". – Sassatelli Giulio Jan 27 '25 at 07:46
  • Interestingly, the set of neighborhoods of $x,$ for fixed $x,$ is itself a topology, and it has a usage. If have a topology $\tau$ on $X,$ and $x\in X,$ we get a topology of neighborhoods, $\tau_x$ (adding in the empty set) and the concept of a function $f:X\to Y$ being "continuous at $x$" is actually just continuous from $(X,\tau_x)\to Y.$ And you get,that continuity on $(X,\tau)$ is equivalent to being continuous at every point of $x.$ – Thomas Andrews Jan 27 '25 at 07:51
  • Even weirder, there is a notion of continuity "to $x.$" There, the topology $\tau^x,$ is just the open sets containing $x,$ which is your second definition, plus the empty set. And a function $Y\to (X,\tau)$ is continuous iff for each $x$, it is continuous to $(X,\tau^x).$ I have no real intuition for what that means, but it is easy to prove. – Thomas Andrews Jan 27 '25 at 07:56
  • I probably shouldn't have tried to type up my correction, still struggling with my new i[ad'a wider keyboard, and getting ahead of myself. – Thomas Andrews Jan 27 '25 at 07:58
  • So, the two different definitions are related to whether you are think of $X$ as the domain or co-domain of functions. – Thomas Andrews Jan 27 '25 at 08:00
  • I don't have a name for these topologies $\tau_x,\tau^x.$ And I got part of the definition of $\tau_x$ wrong - it isn't just your neighborhoods. I came up with them for this answer. https://math.stackexchange.com/a/439481/7933 – Thomas Andrews Jan 27 '25 at 08:04
  • Of possible interest for other characterizations: Reference for the equivalence of alternate formulations of a topological space – Dave L. Renfro Jan 27 '25 at 09:28
  • (4) is more clearly stated as: if $N$ is a neighborhood of $x$, then the set ${y \in N : N\text{ is a neighborhood of }y}$ is also a neighborhood of $x$. (Essentially, "$x$ is in the interior of all its neighborhoods.") (4) is not necessary for uniquely defining the open sets, but if it is not satisfied, then when you redefine neighborhoods from those open sets, you don't get exactly the same collections of neighborhoods. – Paul Sinclair Jan 29 '25 at 22:47
  • Defining topologies by open sets is clearly more concise than neighborhoods. There are four axioms either way, but for open sets, two are trivial, one is pretty much the same as one of yours, and the other is roughly similar to another. None of them are nearly as convoluted as your (4). There are other rankings of definitions than "conciseness", and perhaps topology-by-neighborhoods can be considered superior in some of them (though they tend to be more subjective). – Paul Sinclair Jan 29 '25 at 22:55

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