Must every point have its own singleton as a neighborhood? Edit: For clarity, I provide the link to the particular definition of a topology that I am using.
I am just beginning to study topology, and I had a doubt while going over the 4 axiom definition of neighborhood topology (note: I am avoiding all referencing to open sets here, as I'm talking about the axiomatization using neighborhoods only) - is it necessary for a neighborhood topology to assign to each point a neighborhood that contains the singleton of that point, i.e. for a set $X$, is it true that every $x\in X$ has $\{x\}$ as a neighborhood?
I have come up with the following counterexample: let $X = \{1,2,3\}$, and let the neighborhood topology $\mathcal N (x)$ map the element $1$ to $\mathcal N (1) = \{\{1,2\},\{1,2,3\}\}$. In this case, we do not have any neighborhood only containing $1$. Does this work? I think it probably violates the 4th axiom of the neighborhood topology definition (i.e., any neighbourhood $N$ of $x$ includes a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$), but I am not sure.
 A: There is exactly one topology on a set $X$ that assigns to every $x \in X$ its singleton $\{x\}$ as a neighborhood. That topology is the discrete topology which is just the topology where all the subsets of $X$ are open (in terms of your definition, a subset $U \subseteq X$ is open if it is a neighborhood of every point in $U$).
To see this, let $u \in U$ be any arbitrary point in any arbitrary subset $U \subseteq X$. By the condition above, $\{u\}$ must be a neighborhood of $u$. But obviously $\{u\} \subseteq U$ i.e. $U$ contains a neighborhood of $u$. So, by the second condition of your definition of a topology, $U$ must be a neighborhood of $u$ also.
Thus, $U$ is a neighborhood of every point in itself. Hence, it is open by the definition of openness according to neighborhoods. Since $U$ was an arbitrary subset, we have shown that every subset is open in this topology.
A: To formalize your example:
$$N(1) = (\{1,2\}, \{1,2,3\})$$
$$N(2) = (\{1,2\}, \{1,2,3\})$$
$$N(3) = (\{1,2,3\})$$
Checking the axioms from the linked article:


*

*Is clear. $a$ is in every member of $N(a)$.

*The sets $N$ are $\{1,2\}$ and $\{1,2,3\}$, with the first being a subset of the second. The only proper superset of any $N$ is $\{1,2,3\}$, and this appears as a neighborhood of all points (and in particular those with neighborhood $\{1,2\}$).

*For any $a$ we see the intersection of any two sets in $N(a)$ gives the smaller set itself, which is in $N(a)$.

*In this case, this condition is satisfied by $N$ itself. For example, the set $N = M =\{1,2\}$ is a neighborhood of both $1$ and $2$, the elements of $M$.


And so we've formed a neighborhood topology as desired. Note that no point has a singleton neighborhood.
