How does on prove that
if $(A, \le)$ is a countable upper-directed set, there exist an increasing cofinal sequence $(a_n)_{n \ge 0} \subseteq A$ (i.e. a sequence such that $a_n \le a_{n+1} \ \forall n$ and for any $a \in A$ there exist $n \in \Bbb N$ with $a \le a_n$)?
I am studying Bourbaki's presentation of Kolmogorov's theorem about the existence of the projective limit of a projective system (indexed by $A$) of spaces with measure. The theorem is proven first for $A = \Bbb N$, and for a general countable index set $A$ the proof is reduced to the case $A = \Bbb N$ with the aid of the above statement. No reference or proof suggestion is given therein, and a quick browsing through the set theory volume (the chapter devoted to ordered sets) didn't help me.
Since $A$ is countable, we may write $A=\{b_i: i\in\mathbb{N}\}$. Now we'll build our cofinal sequence in stages:
$a_0=b_0$.
Having defined $a_n$, we let $i$ be the least natural number such that $b_i\ge a_n$ and $b_i\ge b_n$; now set $a_{n+1}=b_i$.
Do you see why this is in fact a cofinal sequence? (HINT: you need to show two things - that $a_0\le a_1\le a_2\le . . . $, and that for each $b\in A$, there is some $a_i\ge b$. Note the two clauses in the definition of $a_{n+1}$; the first is for the first requirement, and the second is for the second requirement.)