I am studying filters and seeing a lot of definitions at once. So I would like to ask you help me.
Is my intuition correct for these guys? Suppose we have partially ordered set with some subsets:
Filter (or order filter) - contains subsets "large enough" to contain ceratin thing - Im calling it just it. Filter is like a way to look for something within given set and tell where it can be located. Empty set is never in the filter (not large enough, useless to look for it there ).For any set in the filter, any bigger sets (containing the set) are also in the filter. Because if a set contains it, also bigger sets contain it of course. If two sets contain it =are in the filter), then also their intersection contains it - is also in the filter.
Ultrafilter - the perfect locating scheme. The best way to look for it. Ultrafilter uses ALL subsets of our given set to determine whether it is hidden in that particular set, or not. From this interpretation, compactness (see the mathematical characterization below) can be viewed as the property that "no location scheme can end up with nothing", or, to put it another way, "always something will be found".)1
Maximal filter - couldn´t find any good definition, could you help? On Wikipedia, it appears to coincide with ultrafilter.
Proper filter - is not equal to the whole set
Principal filter - we take one set and all the sets that are "above" make the principal filter.
z-filter - didn´t find anything about this, I only know that this is used for Stone-Čech compactification when the space is not discrete.
Does this make sense? What else would you add to connect these definitions or put them into context? Any insights welcome.
Note that a more general notion of filter is defined on a poset $(P, \le)$ that has meets (two elements $p,q\in P$ have a greatest lower bound $p \land q \in P$; this is called a meet-semilattice) and a minimum $0 \in P$, and then the axioms for a filter $\emptyset \neq F \subseteq P$ become
So an upwards-closed subset of $P$ (condition 3) that is closed under finite meets (2) and is not equal to $P$ itself (the proper part; note that condition (1) is then needed or otherwise $0 \le x$ for all $x$ and all $x$ would be in $F$, which we don't want).
In a powerset of a set $X$ with $\cap$ as the meet and $\subseteq$ as the poset-order and $\emptyset$ as the minimum, this is precisely your definition as a special case. But in topology we also want to talk about filters of closed sets or zero-sets or open sets and open sets/closed subsets/ zero-sets form a meet semi-lattice too, so there this filter notion also makes sense (we can go further and demand just a common lower bound to be in $F$; this is what Wikipedia does, and is also fine).
An ultrafilter $F$ is just a maximal filter on $P$ so if $P \supseteq F' \supseteq F$ is a filter too, then $F=F'$. For the power set $\mathscr{P}(X)$ meet-semilattice (in fact in any Boolean algebra) it is not hard to see that in fact this is equivalent to
$$\forall A \subseteq X: A \in F \lor X\setminus A \in F$$
so it's then not the case that the ultrafilter is all subsets, but exactly "half" of them (a set and its complement cannot both be in $F$ but precisely one of them has to be).
In this poset (and many others that also consist of subsets of some set $X$), a principal filter is one where the intersection of its members is non-empty, so they have at least one common element. In the power set of $X$, if $F$ is a principal ultrafilter it is of the form
$$F_x=\{A \subseteq X\mid x \in A\}$$ for some $x \in X$, as can easily be seen. Such a filter is always maximal and hence an ultrafilter.
The most commonly considered filter on a topological space $X$ is, for any $x \in X$, the set of neighbourhoods of $x$, denoted $\mathcal{N}_x$, which consists of all subsets $N$ of $X$ so that there is an open subset $O$ of $X$, so that $x \in O \subseteq N$. This is clearly a filter (and a principal one, as all members contain $x$; but not any set containing $x$ is in $\mathcal{N}_x$ in general), and the idea of filter convergence is that we define it so that $\mathcal{N}_x \to x$, the neighbourhoods of a point "converge" to that point.