Integer sequences that satisfy the finite intersection property I am curious what integer sequences are there that satisfy the "finite intersection property". To be precise, what are the family of integer sequences that any intersection of two such sequences in the family (and thus finitely many) is still in the family (or possibly the empty set)?
The only interesting one I know of is the family of arithmetic progressions. Any two arithmetic progressions intersect to be an arithmetic progression or the empty set.
Thanks!
 A: Here is another example. A set $A$ of natural numbers is (temporarily) called interesting if $A$ is the set of perfect $k$-th powers for some positive integer $k$. 
Then the intersection of two interesting sets is interesting. For if $A$ is the set of $m$-th powers, and $B$ is the set of $n$-th powers, then $A\cap B$ is the set of $L$-th powers, where $L$ is the lcm of $m$ and $n$.
Of course the family of finite sets has the required property. So has the family of cofinite sets (sets whose complement is finite).
You may be interested in looking up the relevant term filter. The ultrafilters on $\mathbb{N}$ are of great theoretical interest.
A: Here's a potentially interesting example.  Pick one sequence, $f$, in $\Bbb{Z}$.  Define a subbasis for a topology on $\Bbb{Z}$ by selecting a collection $S$ of sequences where for all $s \in S$, $f \subset s$.  (That is every sequence in $S$ contains $f$ as a subsequence.)  The topology generated by $S$ has the desired property.  (Note that I make no representation this this family of sequences is "small".)
We need not include unions in the above example.  We could arrange for the elements of $S$ to be maximal by instead generating all possible (countable) intersections of them.
We can generalize the above by allowing $F$ to be a set of sequences satisfying the "finite intersection property" and for each $s \in S$, there is an $f \in F$ such that $f \subset s$.  Then generate by countable intersections or generate as a topology.  
Here's a "boring" example that cannot be decomposed recursively into a sequence of $((F_j,S_j))_{j=0}^k$ pairs with $F_j = \langle S_{j-1} \rangle$ and arbitrary $F_0 \neq \varnothing$, so the above constructions cannot exhaust the space of "finite intersection property" families:


*

*Let $s_1 = (1, 3, 7, 15, ... )$, where each term, $t_k$, satisfies $t_k = 2t_{k-1}+1$.

*Let $s_2 = (2, 5, 11, 23, ...)$, where the terms have the same relationship as in $s_1$ and the first term is the first positive integer nowhere in $s_1$.

*Let $s_3 = (4, 9, 19, 39, ...)$ where the first term is the first positive integer appearing nowhere in $s_1$ or $s_2$ and the terms have the same relationship as in $s_1$.

*... and we continue, making each sequence start at the first unused positive integer and have its terms be in the same relationship as those in $s_1$.


If $i \neq j$, $s_i$ and $s_j$ have no term in common.  (Suppose they did.  Since separately, their terms have the same relationships, the pair of terms prior to the common terms are also common.  Inductively, we may march backwards through the two sequences, finding common terms until we come to the first term of one of the sequences.  Either we have come to the first term of both sequences (i.e., $i = j$) or the first term of one sequence appears in some other sequence, contradicting that each sequence starts at a number that has been used by no prior sequence.)
There is a technical requirement:  We need to be sure that we can always make the "next" sequence.  However, in a sequence, the gaps double in size, so after $k$ sequences, there are only $k$ numbers used in the interval from $2^k$ to $2^{k+1}$, which contains many more than $k$ numbers, so there is always a number available with which to start a "next" sequence.
