Suppose that $X$ is a topological space that has been given the discrete topology. Let $p \in X$.
What are the necessary and sufficient conditions for a sequence of points in $X$ to converge to $p$?
What are the necessary and sufficient conditions for a filter on $X$ to converge to p?
solution:
Suppose that $X$ is a topological space with discrete topology, $p \in X$. Let $(x_n)_{n=1}^\infty$ be a sequence in $X$, then $x_n \to p$ iff for some $N, all \\ n \ge N, x_n = p$.
$\Rightarrow$ Suppose $x_n \to p$ and suppose for contradiction there are infinitely many $x_n \neq p$. So there exists a sequence $n_1 < n_2 < n_3 < \dots$ so that $x_ {n_k} \neq p$
Let $U = X / \{x_{n_1}, x_{n_2}, \dots \}$. So $U$ is open in the discrete topology on $X, p \in U$ and there is no $N$ so that for all $n \ge N, x_n \in U$. So $x_n \not\to p$.
$\Leftarrow$ Suppose for some $N_0$, for all $n \ge N_0, \ x_n = p$ then for any set $U$ with $p \in U$, $x_n \in U$ for all $n \ge N_0$. So $x_n \to p$.
A filter $\mathscr F$ on $X$ converges to $p$ when $\mathscr F$ contains every neighborhood of $p$. In the discrete topology, every set $U$ that contains $p$ is a neighborhood of $p$. Claim: $\mathscr F \to p \iff \mathscr F = \{U \subset X: p \in U \}$.
Suppose $\mathscr F \to p$ then for every $U \subset X$, with $p \in U$ is an element of $\mathscr F$. Suppose, however, there exists $V \in \mathscr F, \ p \not\in V$. Since $\{p \} \in \mathscr F$ and $V \bigcap \{p \} = \emptyset$, this then implies $\emptyset \in \mathscr F$, which is not possible. Therefore we have shown that every element $V$ of $\mathscr F$ contains $p$.
$\Leftarrow$ If $\mathscr F = \{U \subset X: p \in U \}$ then $\mathscr F$ is a filter and $\mathscr F \to p$