Lower and upper limits of a filter base

olivierd

Apr 27, 2023 8 min read

Lower and upper limits of a filter base

Two cases, that I have been unable to unify: a filter base over a power set $\mathcal P(X)$ with $X$ a set, or over $\overline \R$ .

First case: filter bases over a power set

$\mathcal B$ filter base over $\mathcal P(X)$ , with $X$ a set.

Definitions

$\mathcal B^* = \bigcap_{\alpha \in \mathcal B} \bigcup_{E \in \alpha} E$

$\mathcal B_* = \bigcup_{\alpha \in \mathcal B} \bigcap_{E \in \alpha} E$

Property

It is always the case that $\mathcal B_* \subseteq \mathcal B^*$ .

Proof: Let $x$ be an eventual element of $\mathcal B_*$ , union for all $\alpha \in \mathcal B$ of the sets $\bigcap_{E \in \alpha} E$ .

Since $x$ is in this union of sets, it is in at least one of them: there exists an $\alpha_0 \in \mathcal B$ such that $x \in \bigcap_{E \in \alpha_0} E$ , that is, such that for all $E \in \alpha_0$ , $x$ belongs to $E$ .

Now let $\alpha$ be any element of $\mathcal B$ . Since $\alpha$ and $\alpha_0$ are both in the filter base $\mathcal B$ , their intersection is non-empty and there exists an $E_1$ in this intersection.

Since $E_1 \in \alpha_0$ , we have seen that $x \in E_1$ .

Furthermore, $E_1 \in \alpha$ .

Hence $x \in \bigcup_{E \in \alpha} E$ .

This being the case for any $\alpha \in \mathcal B$ , we have $x \in \bigcap_{\alpha \in \mathcal B} \bigcup_{E \in \alpha} E$ , that is, $x \in \mathcal B^*$ .

This being the case for all $x \in \mathcal B_$ , we have $\mathcal B_* \subseteq \mathcal B^*$ .

Property

If $\mathcal B$ and $\mathcal B'$ are filter bases over a same $\mathcal P(X)$ , with $\mathcal B'$ finer than $\mathcal B$ , then

$\mathcal B_* \subseteq \mathcal {B'}_*$

$\mathcal B^* \supseteq \mathcal {B'}^*$

That is, the lower limits increase with the fineness of the filter base, and the upper limits decrease.

Proof: $\mathcal B'$ is finer than $\mathcal B$ , that is, every $\alpha \in \mathcal B$ is a superset of some $\alpha' \in \mathcal B'$ .

Let $x$ be an eventual element of $\mathcal B_*$ , that is: $x \in \bigcup_{\alpha \in \mathcal B} \bigcap_{E \in \alpha} E$ . Then there is some $\alpha_1 \in \mathcal B$ such that $x$ is in every $E \in \alpha_1$ . Since $\mathcal B'$ is finer than $\mathcal B$ , $\alpha_1$ is a superset of some $\alpha_2 \in \mathcal B'$ . Then $x$ is in every $E \in \alpha_2$ , that is, $x \in \bigcap_{E \in \alpha_2} E$ , with $\alpha_2 \in \mathcal B'$ . Hence $x \in \bigcup_{\alpha \in \mathcal B'} \bigcap_{E \in \alpha} E = \mathcal {B'}_*$ .

This being the case for all $x \in \mathcal B_*$ , we have $\mathcal B_* \subseteq \mathcal {B'}_*$ .

Let $x$ be an eventual element of $\mathcal {B'}^*$ , that is, $x \in \bigcap_{\alpha \in \mathcal B'} \bigcup_{E \in \alpha} E$ . Let $\alpha_1$ be an element of $\mathcal B$ . Since $\mathcal B'$ is finer than $\mathcal B$ , $\alpha_1$ is a superset of some $\alpha_2 \in \mathcal B'$ . Since $\alpha_2 \in \mathcal B'$ , we have $x \in \bigcup_{E \in \alpha_2} E$ , that is, there is an element $E_1 \in \alpha_2$ such that $x \in E_1$ . Since $\alpha_1 \supseteq \alpha_2$ , we also have $E_1 \in \alpha_1$ , and hence $x \in \bigcup_{E \in \alpha_1} E$ . This being the case for any element $\alpha_1$ of $\mathcal B$ , we have $x \in \bigcap_{\alpha \in \mathcal B} \bigcup_{E \in \alpha} E = \mathcal B^*$ .

This being the case for all $x \in \mathcal {B'}_*$ , we have $\mathcal B^* \supseteq \mathcal {B'}^*$ .

Property

If two filter bases are equivalent, they have the same upper limits and lower limits.

Proof: If the filter bases are equivalent, they are each finer than the other, hence the lower limit of one is both subset and superset of that of the other, that is, are equal; the same for the upper limits.

Sequences of sets

Definitions and expressions of the upper and lower limits of a sequence of sets

Let $E = (E_n)_{n \in \N}$ be a sequence of sets, all subsets of a set $X$ .

The upper and lower limits $E^*$ and $E_*$ of this sequence are classically defined as:

$E^*$ is the set of elements that are in $E_n$ for an infinity of values of $n$ ;

$E_*$ is the set of elements that are in $E_n$ for all but a finite number of $n \in \N$ .

If $x \in E^*$ , then for any $n$ , there is a $p \ge n$ such that $x \in E_p$ , that is, $x \in \bigcup_{p \ge n} E_p$ . Conversely, if for any $n$ we have $x \in \bigcup_{p \ge n} E_p$ , then $x \in E^*$ . Hence:

$E^* = \bigcap_n \bigcup_{p \ge n} E_p$

If $x \in E_*$ , there must be an $n$ such that all the values of $p$ for which $x \notin E_p$ are less than $n$ ; then for all $p \ge n$ , we have $x \in E_p$ , that is, $x \in \bigcap_{p \ge n} E_p$ . Conversely, if there is an $n$ such that $x \in \bigcap_{p \ge n} E_p$ , then all values of $p$ such that $x \notin E_p$ must be less than $n$ , and hence there are a finite number of them; thus $x \in E_*$ . Hence:

$E_* = \bigcup_n \bigcap_{p \ge n} E_p$

Fréchet filter and filter base

The Fréchet filter $\mathcal F_{\text F}$ is the filter on $\N$ the elements of which are the cofinite subsets of $\N$ . Let $\mathcal B_{\text F}$ be defined as the set of subsets $\N_n = \{p \in \N\ |\ p \ge n\}$ for all $n \in \N$ . It is easily checked that $\mathcal B_{\text F}$ is a filter base and generates $\mathcal F$ on $\N$ ; we will call it the Fréchet filter base.

Fréchet filter and upper/lower limits of a sequence of sets

Again, let $E = (E_n)_{n \in \N}$ be a sequence of subsets of a set $X$ .

We can consider $E$ as a mapping $\N \to \mathcal P(X)$ ; the Fréchet filter base $\mathcal B_{\text F}$ has a certain image $\mathcal B$ by this mapping, that is a filter base on $\mathcal P(X)$ . An element $\N_n$ of $\mathcal B_{\text F}$ corresponds to the element $E^\rightarrow(\N_n) = \{E_p\}_{p \ge n}$ .