# Filters and sequences

Let $X$ be a set, let $(x_n)$ and $(y_n)$ be sequences in $X$ and let $\mathcal F$ and $\mathcal G$ be the filters generated by the tails of $(x_n)$ and $(y_n)$ respectively. Is there any interpretation of $\mathcal F \cap \mathcal G$ and the least upper bound of $\mathcal F$ and $\mathcal G$ (when it exists) in terms of the sequences $(x_n)$ and $(y_n)$?

It seems to me that if we interleave the sequences $(x_n)$ and $(y_n)$ into a new sequence $(z_n)$, the filter $\mathcal H$ generated by the tails of $(z_n)$ is always finer than $\mathcal F \cap \mathcal G$. I guess this means that $\mathcal F \cap \mathcal G$ is the intersection of all filters generated by the tails of sequences that are made using $(x_n)$ and $(y_n)$ through interleaving.

I have no clue about least upper bound of $\mathcal F$ and $\mathcal G$. Maybe it has something to do with a subsequence of $(x_n)$ related to a subsequence of $(y_n)$.

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Let $S_n=\{x_k:k\ge n\}$ and $T_n=\{y_k:k\ge n\}$. Let $A\subseteq X$; then $A\in\mathscr{F}\cap\mathscr{G}$ iff $S_n\cup T_n\subseteq A$ for some $n\in\Bbb N$. One direction is obvious. For the other, suppose that $A\in\mathscr{F}\cap\mathscr{G}$; then there are $m,n\in\Bbb N$ such that $S_m\cup T_n\subseteq A$. Let $r=\max\{m,n\}$; then $S_r\cup T_r\subseteq S_m\cup T_n\subseteq A$.

Now let $z_{2n}=x_n$ and $z_{2n+1}=y_n$ for $n\in\Bbb N$, one of the sequences obtained by interleaving the sequences $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$, and for $n\in\Bbb N$ let $U_n=\{z_k:k\ge n\}$; it follows from the previous paragraph that $A\in\mathscr{F}\cap\mathscr{G}$ iff $U_n\subseteq A$ for some $n\in\Bbb N$ and hence that $\mathscr{F}\cap\mathscr{G}$ is the filter generated by $\{U_n:n\in\Bbb N\}$. In other words, you need only look at this one interleaved sequence. (With a bit more work you can show that any interleaved sequence, even an irregular one, will work.)

I assume that by the least upper bound of $\mathscr{F}$ and $\mathscr{G}$ you mean the intersection of all filters containing $\mathscr{F}\cup\mathscr{G}$, provided that there are any. Let $\mathscr{H}$ be this filter, assuming that it exists. For each $F\in\mathscr{F}$ and $G\in\mathscr{G}$ we have $F,G\in\mathscr{H}$, so $F\cap G\ne\varnothing$. In particular, for all $m,n\in\Bbb N$ we must have $S_m\cap T_n\ne\varnothing$. This is equivalent to the following assertion:

$\qquad\quad$ for each $m,n\in\Bbb N$ there are $k,\ell\in\Bbb N$ such that $k\ge m$, $\ell\ge n$, and $x_k=y_\ell$.

A little thought shows that this is equivalent to the following slightly simpler version:

$\qquad\qquad\qquad\quad$ for each $m\in\Bbb N$ there are $k,\ell\ge m$ such that $x_k=y_\ell$.

It’s easy to see that the condition that $S_m\cap T_n\ne\varnothing$ for all $m,n\in\Bbb N$ is not just necessary but also sufficient for $\mathscr{F}$ and $\mathscr{G}$ to have a least upper bound. Thus, $\mathscr{F}$ and $\mathscr{G}$ have a least upper bound iff the sequences $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ have infinitely many terms in common (not necessarily with the same indices). But that just says that they have a common subsequence. Thus, we can finally say that $\mathscr{F}$ and $\mathscr{G}$ have a least upper bound iff the sequences $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ have a common subsequence.

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