The Elementary Filter Associated with a Sequence

# The Elementary Filter Associated with a Sequence

Let $E$ be a topological space and let $(x_n) \subseteq E$. For each $n \in \mathbb{N}$ let:

(1)\begin{align} \quad X_n := \{ x_i : i \geq n \} \end{align}

Then observe that $\{ X_n : n \in \mathbb{N} \}$ is a filter base. Indeed if $m, n \in \mathbb{N}$ and $m \leq n$ then $X_n$ is contained in $X_m \cap X_n$. The filter generated by $\{ X_n : n \in \mathbb{N} \}$ is given a special name.

Definition: Let $E$ be a topological space and let $(x_n) \subseteq E$. The Elementary Filter Associated with $(x_n)$ is the filter generated by the sets $\{ X_n : n \in \mathbb{N} \}$ where $X_n := \{ x_i : i \geq n \}$. |

*So the elementary filter associated with $(x_n)$ is the collection of all subsets of $E$ which contain a tail of the sequence.*

Proposition 1: Let $E$ be a topological space and let $(x_n) \subset E$. Then $(x_n)$ converges to $a$ if and only if the elementary filter associated with $(x_n)$ converges to $a$. |

**Proof:**$\Rightarrow$ Suppose that $(x_n)$ converges to $a$. Let $U$ be a neighbourhood of $a$. Then there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $x_n \in U$. So $X_N \subseteq U$. But by definition, the elementary filter associated with $(x_n)$ contains all sets which contain a set in $\{ X_n : n \in \mathbb{N} \}$ so $U$ is in the elementary filter associated with $(x_n)$. Hence the elementary filter associated with $(x_n)$ is a refinement of the filter consisting of neighbourhoods of $a$ and so by one of the propositions on the Convergence of Filters and Filter Bases in a Topological Space page we have that the elementary filter associated with $(x_n)$ converges to $a$.

- $\Leftarrow$ Suppose that the elementary filter associated with $(x_n)$ converges to $a$. Let $U$ be a neighbourhood of $a$. Then $U$ contains a set in $\{ X_n : n \in \mathbb{N}$, i.e., there exists an $N \in \mathbb{N}$ such that $X_N \subseteq U$. Thus $x_n \in U$ if $n \geq N$, so $(x_n)$ converges to $a$. $\blacksquare$