Let $\mathscr F$ be the collection of subsets of $[0,1]$ which contain an interval $[\frac{1}{3},\frac{2}{3}]$. Use the relative topology on $[0,1]$ considered as a subspace of $\mathbb R$ with the standard metric topology. Prove:
- $\mathscr F$ is a filter
- $\mathscr F$ does not converge
- Find filter $\mathscr G$ on $[0,1]$ so if $F \in \mathscr F$ then $F \in \mathscr G$, but $\mathscr G$ converges.
Solution: Let $\mathscr F = \{U \subset [0,1]: [\frac{1}{3},\frac{2}{3}] \subset U \}$.
$\mathscr F$ is a filter: $\emptyset \notin \mathscr F$. Suppose $U_1, U_2 \in \mathscr F$, then $[\frac{1}{3},\frac{2}{3}] \subset U_1 \bigcap U_2$, so $ U_1 \bigcap U_2 \in \mathscr F$. Suppose $U \in \mathscr F$ and there is a set $S$ such that $U \subset S$, then $[\frac{1}{3},\frac{2}{3}] \subset U$, so $[\frac{1}{3},\frac{2}{3}] \subset S$, thus $S \in \mathscr F$
$\mathscr F$ does not converge. Suppose for the sake of contradiction that for some point $p \in [0,1]$, $\mathscr F \to p$.
If $p > \frac{2}{3}$ or $p < \frac{1}{3}$ there is an open neighborhood $V$ of $p$ so that $V \bigcap [\frac{1}{3},\frac{2}{3}] = \emptyset$. Since $\emptyset \notin \mathscr F$, $\mathscr F \not\to p$.
If $\frac{1}{3} < p< \frac{2}{3}$, then there exists $\epsilon >0$ so that $(p-\epsilon, p+\epsilon) \subset (\frac{1}{3},\frac{2}{3})$. Since $(p-\epsilon, p+\epsilon)$ is a neighborhood of $p$ and not contained in $\mathscr F$, $\mathscr F \not\to p$.
If $p=\frac{1}{3}$ or $p=\frac{2}{3}$, consider $V=(p- \frac{1}{6}, p+ \frac{1}{6} )$, $V$ is a neighborhood of $p$, but $V \bigcap [\frac{1}{3},\frac{2}{3}]$ does not contain $[\frac{1}{3},\frac{2}{3}]$ so $V \notin \mathscr F$ and $\mathscr F \not\to p$.
- Let $\mathscr G = \{U \subset [0,1]: \frac{1}{3} \in U \}$. then every element $U$ of $\mathscr F$ contains $[\frac{1}{3}, \frac{2}{3}]$ and so $\frac{1}{3} \in U$ hence $U \in \mathscr G$. Also $\mathscr G$ is a filter and $\mathscr G \to \frac{1}{3}$