Is the definition of a filter a circular definition?
I saw the definition of a filter for the first time and am confused. Filter is supposed to be a collection F of subsets of a set X with the following three properties:
1. Empty set is not in F. No problem with this.
2. Any finite intersection of sets in F belongs to F. Also no problem.
3. Any subset of X containing a set of F belongs to F.
The #3 seems circular to me. If we start with nothing, how do we know which subset contains a set of F? I understand, that inuitively, the filter describes sets "large enough" to do something.
I have heard about principal filters (all subsets of X that contain given point) or neighborhood filters (all subsets of X having given point as their inferior point). These definitions seem okay to me, since we have something to start with. But generally, I don´t see it that way.
Where do we start when "choosing" the sets of the filter? Given a set X, how do we choose some sets belonging to F? Or does filter definition always come with further specification? Could you provide some examples?
Thank you in advance.
$\endgroup$ 22 Answers
$\begingroup$This would be a problem if the author was trying to define a specific filter, but this is not what is being done.
The author is defining what a filter is in general.
Once you have a specific filter you should be able to verify if these three axioms are satisfied.
Let's work with the most common filter I have seen:
$X = \mathbb R$ and $F=\{ A \subseteq \mathbb R | 0 \in A\}$.
$1$. The empty set is not in $F$ as $0 \not \in \varnothing$.
$2$. If $A\in F$ and $B\in F$ then $0\in A$ and $0\in B$ and so $0\in A\cap B$ and so $A\cap B \in F$.
$3$. if $A\in F$ and $S \supseteq A$ we have $0\in S$ and so $S\in F$.
$\endgroup$ 2 $\begingroup$The definition isn't circular. Perhaps a more intuitive way of stating #3 is
3.) F is closed under the operation of taking supersets.
This way of rewording #3 gives rise to some immediate examples. Let $N\subset X$. Then let
$$\mathcal{F}_N=\{K\subset X: N\subset K\}.$$
So a $K\in\mathcal{F}_N$ if $N$ is a subset of $K$. You can check that $\mathcal{F}_N$ is a filter on $X$.
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