Which filters converge in a discrete topological space?
Is this correct:
Let $(X,\tau)$ discrete topological space and $\cal{F}$ a filter and $\cal{U}_x$ the neighborhood system of $x$. Then $\{x\}$ is a neighborhood of $x$. By definition, $\cal{F} \to x \iff \cal{U}_x \subseteq \cal{F}$
Since every neighborhood of $x$ must be in the filter and $\{x\}$ is a basic neighborhood, the only convergent filter is $\cal{F} = \cal{U}_x$ (Is this statement correct?)
Thanks.
Best Answer
And so $\mathcal{F} \to x$ implies that $\mathcal{F} = \{A \subseteq X: x \in A\}$, the fixed filter on $X$. So there is exactly one filter converging to $x$, for every $x \in X$. This is due to $\{x\} \subseteq A$ iff $x \in A$ and filters being closed under supersets, plus $\mathcal{F} \to x$ iff $\{x\} \in \mathcal{F}$.