Definition of elementary filters and their existences for certain sequences.

Question

I have read, from the book $ \textit{Topological and Uniform Spaces} $, the following definition of an $ \textit{elementary filter} $:

Let $ (x_n ) $ be an sequence of points of the set $X$. The elementary filter associated with $(x_n)$ is the filter consisting of the subsets $M$ of $X$ for which there exists an integer $k$ such that $ x_n \in M $ whenever $ n \geq k $.

However I find this definition to be ambiguous in a number of ways.

1. Is the $k$ uniquely determined by the elementary filter? Or is it possible for us to have many choices of $k$?

2. If an elementary filters exists, is it unique? As an example, the book offers the sequence $ (n) $ with the nth element being $n \in \mathbb{N} $, and claims that the elementary filter associated with this sequence consists of the cofinite subsets of the set $\mathbb{N}$. However in this case, if we define cofinites subsets to be the set of complements of finite subsets. Then for every $k$ and every $ n \geq k $, if we consider the set $ \mathbb{N} \backslash \{ n \} $, this is a complement of a finite set so it is a cofinite set, but $ n \notin \mathbb{N}\backslash\{n\} $. So there exists $n$ for which $x_n$ is not in this elementary filter.

What am I thinking wrong here???

Answer 1

1. The definition says "there exists some $k$". It cannot be unique: if $k$ works for $M$, also $k+1$ works for $M$. You could consider the smallest $k$ that works for a single $M$, but the whole filter does not need to have a common $k$ that works for all the sets in it (see 2.).
2. The filter associated to a sequence is unique, because its definition is not ambiguous: it consists of all the sets that contain all elements of the sequence except possibly finitely many (this is exactly equivalent to saying "there exists $k$...", can you see why?). Given any set $M$, you can determine whether it belongs to the filter of $(x_n)$ or not: it belongs to the filter if and only if $M \cap (x_n)$ has infinitely many elements. In your example, asking for a set $M$ to be in the filter of the sequence $(n)$ is equivalent to asking that there is some $k$ such that all the $n \ge k$ are in $M.$ Equivalently, $M$ is a cofinite set, because if $M$ is in the filter, the only elements that could (although they don't need to!) be outside $M$ are the $n <k$ (for some $k$), and these are finitely many. For your specific $M=\mathbb{N} \backslash \{n\}$, just take $k:= n+1$ to see that $M$ is in the filter of $(n)$.