Topology on a 3-elements set: how are the definitions via open set and via neighborhood consitent with each other? 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.
 A: 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).
A: 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.
