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Given the set of 3 points X={A, B, C}, wikipedia says that the collection N={{},{A},{A,B,C}} is a topology on X according to the so-called open-sets definition. I.e., N is a topology because: (i) the empty set and X both belong to N; (ii) any arbitrary union of elements of N belongs to N; (iii) arbitrary intersection of elements of N also belongs to N.

But the definition via neighborhoods says that every superset of a neighborhood of a point in X must also be a neighborhood of this point. So if {A} is a neighborhood of A, then {A,B} and {A,C} should also be a neighborhood of A.

To me it seems that the collection N={{},{A},{A,B,C}} is a topology according to the open-sets definition, but not according to the neighborhoods definition. Could anyone explain this seeming "inconsistency"?

P.s.: I am not a mathematician, so please go easy on terminology.

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  • $\begingroup$ From the same wikipedia page: "Given such a structure, a subset U of X is defined to be open if U is a neighbourhood of all points in U." This means that not every neighborhood is open. $\endgroup$
    – Javi
    Commented Feb 3, 2021 at 11:20

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Indeed for $X=\{a,b,c\}$ the collection $\mathcal{T}=\{\{a\},\{a,b,c\},\emptyset\}$ is a topology that obeys the usual open set axioms.

It's not how a topology via neighbourhoods is defined, for that we have to specify a set of neighbourhoods for each point. For this topology (using that a neighbourhood of $x$ is any set that contains an open set containing $x$) the equivalent description in terms of neighbourhoods per point is

$$\mathcal{N}(a)=\{\{a\},\{a,b\},\{a,c\},\{a,b,c\}\}, \mathcal{N}(b)=\{\{a,b,c\}\}, \mathcal{N}(c)=\{\{a,b,c\}\}$$

but if you define the space that way, you have to check a different set of axioms that are defined in terms of such neighbourhood systems. Now I know these are satisfied already because I start with a standard open-set topology.

Both ways of looking at it give the same notions of open/closed/continuity/convergence etc., so in practice we use whatever description is more convenient.

But do note that the descriptions have different "type": one is a single collection of subsets we call open, the other a function that assigns to each point a collection of subsets (the neighbourhoods of that points).

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    $\begingroup$ In this rather technical answer I discuss in detail how we go from a description in terms of neighbourhoods to the one in terms of open sets and vice versa. In case anyone is interested. $\endgroup$ Commented Feb 3, 2021 at 11:45
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If you read the whole section, before the definition of Topology via open sets you can read the followin about the definition via neighbourhoods

Given such a structure, a subset $U$ of $X$ is defined to be open if $U$ is a neighbourhood of all points in $U$.

In particular, this implies that not every neighbourhood is an open set. It seems to me that you are assumming that a neighbourhood is the same thing a superset, but you actually need to specify what the neighbourhoods of a points are. Neighbourhoods are always supersets of the points the are associated to, but not every superset must be a neighbourhood.

If you take as neighbourhoods of each point all its supersets, you can check that in your example you also get that the subsets $\{A\}$ and $\{A,B,C\}$ are open with respect to the definition of open sets in terms of neighbourhoods given above.

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