Open intervals $(x-r;x+r)$ form a filter base on $\Bbb R$

Question

Let $x \in \Bbb R$ and let $\bf B$ be the collection of open intervals $(x-r; x+r)$ where $r > 0$.

$\bf B$ is a filter base on $\Bbb R$.

Can someone please explain why $\bf B$ is a filter base on $\Bbb R$?

Answer 1

Let $X$ be any non-empty set. A filter on $X$ is a non-empty family $\mathscr{F}$ of subsets of $X$ with the following properties:

1. $\varnothing\notin\mathscr{F}$;
2. if $B\supseteq A\in\mathscr{F}$, then $B\in\mathscr{F}$; and
3. if $A,B\in\mathscr{F}$, then $A\cap B\in\mathscr{F}$.

(2) says that if some subset $A$ of $X$ is in the filter, then so is every subset of $X$ that contains $A$; we sometimes say that $\mathscr{F}$ is closed under taking supersets. (3) implies that $\mathscr{F}$ has the finite intersection property: the intersection of any finite number of members of $\mathscr{F}$ is a member of $\mathscr{F}$. (The proof is an easy induction on the number of sets being intersected.) We can also express this by saying that $\mathscr{F}$ is closed under (taking) finite intersections. (1) just says that $\mathscr{F}$ isn’t the entire power set of $X$.

A collection $\mathscr{B}\subseteq\mathscr{F}$ is a base for the filter $\mathscr{F}$ if closing $\mathscr{B}$ under finite intersections and supersets gives you $\mathscr{F}$. That is, if we let $\mathscr{B}^*$ be the set of all intersections of finite subsets of $\mathscr{B}$, the closure of $\mathscr{B}$ under finite intersections, then

$$\mathscr{F}=\{A\subseteq X:\exists B\in\mathscr{B}^*(B\subseteq A)\}\;.$$

Finally, a non-empty family $\mathscr{B}$ of subsets of $X$ is a filter base on $X$ if it is the base for some filter on $X$. This means that if we form $\mathscr{B}^*$ as above, taking it to be the set of all intersections of finite subfamilies of $\mathscr{B}$, then

$$\{A\subseteq X:\exists B\in\mathscr{B}^*(B\subseteq A)\}$$

is a filter on $X$. It’s not hard to prove that this will be the case if and only if $\mathscr{B}$ satisfies the following two conditions:

1. if $U,V\in\mathscr{B}$, then there is a $B\in\mathscr{B}$ such that $B\subseteq U\cap V$, and
2. $\varnothing\notin\mathscr{B}$.

In your case $\mathscr{B}=\{(x-r,x+r):r>0\}$. Clearly $\varnothing\notin\mathscr{B}$, so condition (2) is satisfied. If $(x-r,x+r),(x-s,x+s)\in\mathscr{B}$, let $t=\min\{r,s\}$; then

$$(x-r,x+r)\cap(x-s,x+s)=(x-t,x+t)\in\mathscr{B}\;,$$

which is actually more than is required by condition (1): $\mathscr{B}$ already has the finite intersection property. It follows that $\mathscr{B}$.

We can even say pretty easily what the filter $\mathscr{F}$ generated by this filter base is. First, since $\mathscr{B}$ is already closed under finite intersections, $\mathscr{B}^*=\mathscr{B}$, and

$$\begin{align*} \mathscr{F}&=\{A\subseteq\Bbb R:\exists F\in\mathscr{F}(F\subseteq A)\}\\ &=\left\{A\subseteq\Bbb R:\exists r>0\Big((x-r,x+r)\subseteq A\Big)\right\}\;. \end{align*}$$

In other words, $\mathscr{F}$ is the family of all subsets of $\Bbb R$ that contain an open interval about the point $x$.