Topological space in which the principal filters are the only filters that converge Let $(X, \mathcal{T})$ be a topological space in which only the principal filters converge. Show that $\mathcal{T}$ is the discrete topology.
It is similar to one of my previous questions (link: Topological space in which every filter in which every filter converges to every point), but it needs a different approach that I can't (yet) come up with.
Definitions used (for the sake of consistency):


*

*A filter is a nonempty set $\mathcal{F}$ for which the following properties hold: $\mathcal{F}$ does not contain the empty set; for every  $F \in \mathcal{F}$ such that $F \subset G, G \in \mathcal{F}$ holds; for every $F \in \mathcal{F}$ and $G \in \mathcal{F} $ also $F \cap G \in \mathcal{F}$.

*A principal filter generated by a set $A \subset X$ is the set$\{F \subset X \vert A \subset F\}$.

*A filter $\mathcal{F}$ converges to $x \in X$ iff the neighbourhood filter of $x$ is contained in $\mathcal{F}$ or, equivalently, for every $V$ in the neighbourhood filter of $x$, there exists an element $F \in \mathcal{F}$ such that $F \subset V$.

*A subset $V \subset X$ is called a neighbourhood of $x$ if there exists an open set $T \in \mathcal{T}$ such that $x \in T \subset V$.

*The neighbourhood filter of $x$ is the set of all neighbourhoods of $x$.

 A: This isn't true; for instance, if $X$ has only finitely many points, then every filter on $X$ is principal, so the condition holds rather trivially for any topology on $X$.
It is true if you assume $X$ is $T_1$.  To show this, note that for any $x\in X$, the neighborhood filter of $x$ converges to $x$ and thus must be principal, generated by some set $A$.  If $A$ contains any point other than $x$, you can now use the $T_1$ hypothesis to get a contradiction.
A: [This is the third version of this answer.  Thanks to Eric Wofsey for corrections on prior versions.]
Consider the following properties on a topological space $X$:  
(i) Every convergent filter on $X$ is principal.
(ii) Every point of $X$ has a finite neighborhood.
(iii) Every point of $X$ has a minimal neighborhood.
(iv) Every neighborhood filter on $X$ is principal.
(v) Arbitrary intersections of open subsets are open.  
I claim (i) $\iff$ (ii) $\implies$ (iii) $\iff$ (iv) $\iff$ (v).  Spaces satisfying the last three properties are called Alexandroff spaces.  I have (evidently!) never met conditions (i) $\iff$ (ii) before.  
Proofs: (iii) $\iff$ (iv) is immediate, since a filter is principal iff it has a minimal element.  (iii) $\iff$ (v) is straightforward (and standard).  (ii) $\implies$ (iii): if $U$ is a finite neighborhood of a point $x$ which is not minimal, then there is a neighborhood $V$ of $x$ not containing $U$ and thus $U \cap V$ is a strictly smaller finite neighborhood.  Repeating this process we get to a minimal neighborhood.  
(i) $\implies$ (ii): If (i) holds, then certainly (iv) holds, hence also (iii) holds.  If for some point $x$ the minimal neighborhood $U_x$ (i.e., the intersection of all neighborhoods of $x$) is infinite, then the collection of subsets of $X$ which contain all but finitely many elements of $U_x$ is a nonprincipal filter converging to $x$.  
(ii) $\implies$ (i): If $\mathcal{F} \rightarrow x$ then $\mathcal{F}$ contains the neighborhood filter at $x$, which is the principal filter associated to a finite set $U_x$.  The filters containing the principal filter associated to a finite set $U_x$ are the principal filters associated to the finite nonempty subsets of $U_x$.  
N.B.: (iii) does not imply (ii): endow any infinite set $X$ with the indiscrete topology.  Then $X$ itself is the minimal neighborhood of all of its points.  Moreover every filter on $X$ converges to every point on $X$, so the Frechet filter is convergent and nonprincipal.
In particular, as Eric points out, every finite space has property (i), and every separated ($T1$) space satisfying property (i) is discrete.  
