Convergence of Filters and Filter Bases in a Topological Space

# Convergence of Filters and Filter Bases in a Topological Space

Definition: Let $E$ be a topological space. If $\mathscr{F}$ is a filter, we say that $\mathscr{F}$ Converges to $a$, and write $\mathscr{F} \to a$, if every neighbourhood of $a$ contains a set in $\mathscr{F}$. |

Definition: Let $E$ be a topological space. If $\mathscr{B}$ is a filter base, we say that $\mathscr{B}$ Converges to $a$, and write $\mathscr{B} \to a$, if every neighbourhood of $a$ contains a set in $\mathscr{B}$. |

We will now discuss some convergence criteria.

Proposition 1: Let $E$ be a topological space and let $\mathscr{F}$ be a filter. Then the following statements are equivalent:(1) $\mathscr{F}$ converges to $a$.(2) Every neighbourhood of $a$ is in $\mathscr{F}$.(3) $\mathscr{F}$ is a refinement of the filter consisting of the neighbourhoods of $a$. |

**Proof:**$(1) \Rightarrow (2)$ Suppose that $\mathscr{F}$ converges to $a$. Let $U$ be a neighbourhood of $a$. Then there exists a set $V \in \mathscr{F}$ such that $V \subseteq U$. But since $\mathscr{F}$ is a filter, $V \in \mathscr{F}$, and $V \subseteq U$, we have by the definition that $U \in \mathscr{F}$. So every neighbourhood of $a$ is in $\mathscr{F}$.

- $(1) \Leftarrow (2)$ Trivial.

- $(1) \Rightarrow (3)$ Suppose that $\mathscr{F}$ converges to $a$. Let $U$ be a neighbourhood of $a$. Then by definition, there exists an $A \in \mathscr{F}$ such that $A \subseteq U$. Since $\mathscr{F}$ is a filter and $A \subseteq U$ we have that $U \in \mathscr{F}$. So every neighbourhood of $a$ is contained in $\mathscr{F}$ so that $\mathscr{F}$ is a refinement of the filter consisting of the neighbourhoods of $a$.

- $(3) \Leftarrow (1)$ Suppose that $\mathscr{F}$ is a refinement of the filter consisting of the neighbourhoods of $a$. Then trivially, any neighbourhood $U$ of $a$ is such that $U \in \mathscr{F}$, so that by definition, $\mathscr{F}$ converges to $a$. $\blacksquare$

Proposition 2: Let $E$ be a topological space and let $\mathscr{F}$ be a filter. If $\mathscr{F}$ converges to $a$ then $a \in \overline{A}$ for every $A \in \mathscr{F}$. |

**Proof:**Suppose that $\mathscr{F}$ converges to $a$ and let $A \in \mathscr{F}$. Let $U$ be a neighbourhood of $a$. Then by Proposition 1 we must have that $U \in \mathscr{F}$.

- Since $\mathscr{F}$ is a filter and since $A, U \in \mathscr{F}$ we have that $A \cap U \in \mathscr{F}$. But $\mathscr{F}$ contains only nonempty sets, and thus $A \cap U \neq \emptyset$. Since $A \cap U \neq \emptyset$ for every neighbourhood $U$ of $a$ and thus $a \in \overline{A}$. $\blacksquare$

Proposition 3: Let $E$ be a Hausdorff topological space. If $\mathscr{F}$ is a filter that converges to $a$ and $b$ then $a = b$. |

**Proof:**Suppose that $\mathscr{F}$ converges to $a$ and $\mathscr{F}$ converges to $b$ and that instead $a \neq b$. Then there exists neighbourhoods $U$ of $a$ and $V$ of $b$ such that:

\begin{align} \quad U \cap V = \emptyset \end{align}

- But since $\mathscr{F}$ converges to $a$ we have by Proposition 1 that $U \in \mathscr{F}$. Similarly, since $\mathscr{F}$ converges to $b$ we have that $V \in \mathscr{F}$. But since $\mathscr{F}$ is a filter and $U, V \in \mathscr{F}$ we have that $U \cap V \in \mathscr{F}$. But $\mathscr{F}$ does not contain the empty set by definition, so we have arrived at a contradiction. Thus we must have that $a = b$. $\blacksquare$