Suppose $\langle L,\leq\rangle$ is a lattice with join $\sqcup$. Let $F\_1$ and $F\_2$ be principal filters on $L$. Thus, for $i\in I=\{1,2\}$ there are $x\_i\in L$ so that $F\_i=\{y\in L:x\_i\leq y\}$.


In this situation, $F\_1\cap F\_2$ is also a principal filter, because $F\_1\cap F\_2=\{y\in L:x\_1\sqcup x\_2\leq y\}$. This isn't too hard to see: clearly if $x\_1\sqcup x\_2\leq y$, then $x\_1\leq y$ (so $y\in F\_1$) and $x\_2\leq y$ (so $y\in F\_2$). Thus $\{y\in L: x\sqcup y\leq y\}\subseteq F\_1\cap F\_2$. On the other hand, if $y\in F\_1\cap F\_2$, then $x\_1\leq y$ (because $y\in F\_1$) and $x\_2\leq y$ (because $y\in F\_2$) and thus $x\_1\sqcup x\_2\leq y$. So $F\_1\cap F\_2\subseteq\{y\in L:x\_1\sqcup x\_2\leq y\}$.


Inductively, if $I$ is finite then $\bigcap\_{i\in I} F\_i$ is a principal filter. But what if $I$ is arbitrary? In general, of course, the intersection of an arbitrary family of principal filters might be empty (example: take the divisibility poset on $\mathbb{N}$, and let $[p]$ be the principal filter generated by a prime $p$. The intersection of all such filter is clearly empty, since no number is divisible by every prime). Ok, sure. But (now the actual question):



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> provided each $F\_i$ is a principal filter and $\bigcap\_{i\in I} F\_i$ is nonempty, is it a principal filter?
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In all the examples I can think of, ensuring the big intersection is nonempty essentially guarantees that the result is a principal filter. So e.g. in the lattice of subsets of $\mathbb{N}$, let $E$ be the set of all subsets of the even numbers, and let $F\_E$ be the set containing, for each $e\in E$, the principal filter generated by $e$. Then, lo and behold, the intersection of all members of $F\_E$ is the filter generated by $E$. And clearly if $L$ is complete, then the answer is an easy 'yes'. So I guess I'm really interested in what we can say *generally*, for any old lattice $L$.



Are arbitrary nonempty intersections of principal filters principal?
Nope. For a silly counterexample consider the lattice of nonzero real numbers with the usual order. Consider the principal filters $F\_i = \left\{y \ne 0: -\frac{1}{i} \le y \right\}$. Then $F=\bigcap\_i F\_i $ should be the principal filter generated by zero. But since zero is not in the lattice $F$ is the filter of positive elements. This is nonprincipal.


