A question about filters. In this article on filters, it says 

For every $x,y\in F$, there is an element $z\in F$ such that $z\leq x$ and $z\leq y$.

How is this different from saying "for any $x\in F$, there is a $z\in F$ such that $z\leq x$"?
Thanks
 A: For one thing, your proposed substitution actually trivially holds for all $F \subseteq P$, since $x \leq x$ is always true (recall that partial orders are reflexive).

Addition:
What I intend to mean by the above is the following: If you instead defined a "filter" to be any $F \subseteq P$ satisfying


*

*for any $x\in F$, there is a $z\in F$ such that $z\leq x$;

*for every $x \in F$ and $y \in P$, $x \leq y$ implies that $y \in F$; and

*$F \neq P$


then the "filters" are just the upwards-closed proper (nonempty) subsets of $P$, since condition (1) is true for all (nonempty) subsets of $P$.
This is a great departure from the usual definition, since now, considering the partial order $\langle \mathcal{P}( \mathbb{N} ) , \subseteq \rangle$, the family $$ F = \{ a \subseteq \mathbb{N} : 0 \in a \} \cup \{ a \subseteq \mathbb{N} : 1 \in a \}$$ is a "filter".  But note that $\{ 0 \}$ and $\{ 1 \}$ are both elements of $F$, and there is no element of $F$ simultaneously below each of these; thus it is not a filter in the usual sense.
A: Saying that you can always find one element $z$ simultaneously satisfying $z\leq x$ and $z\leq y$ is stronger than saying $z\leq x$ for some $z$ and $w\leq y$ for some $w$.
The poset $S=\{w,x,y\}$ with ordering
$$x\leq w,\qquad y\leq w,\qquad x\text{ and }y\text{ are incomparable}$$
satisfies your condition (for the reason Arthur Fischer mentioned), but no element of $S$ is simultaneously $\leq x$ and $\leq y$.
