Cauchy Filters

Definition: Let $E$ be a locally convex topological vector space. A Cauchy Filter is a filter $\mathscr{F}$ with the property that $\mathscr{F}$ contains a set small of order $U$ for each neighbourhood $U$ of the origin.

The following proposition ensures us that every convergent filter is a Cauchy filter.

Proposition 1: Let $E$ be a locally convex topological vector space. If $\mathscr{F}$ is a convergent filter than $\mathscr{F}$ is a Cauchy filter.

• Proof: Suppose that $\mathscr{F}$ converges, say to $a \in E$. Let $U$ be a neighbourhood of the origin. Since $E$ is a locally convex topological vector space there exists an absolutely convex neighbourhood $V$ of the origin such that $V \subseteq U$.

• By the definition of $\mathscr{F}$ converging to $a$, the neighbourhood $a + \frac{1}{2}V \subseteq a + \frac{1}{2} U$ of $a$ is contained in $\mathscr{F}$. But $a + \frac{1}{2}V$ is small of order $U$. Indeed, if $x, y \in a + \frac{1}{2}V$ then $x = a + \frac{1}{2}v_1$ and $y = a + \frac{1}{2}v_2$ for some $v_1, v_2 \in V$. Then:

(1)

\begin{align} \quad x - y = \frac{1}{2}v_1 - \frac{1}{2}v_2 \in \frac{1}{2}V - \frac{1}{2}V = V \subseteq U \end{align}

• (Where we note that $V = -V$ since $V$ is absoluitely convex). Thus $\mathscr{F}$ contains a set small of order $U$ for each neighbourhood $U$ of the origin, and so $\mathscr{F}$ is a Cauchy filter. $\blacksquare$