Existence of a monotone (and cofinal) choice function

Question

Suppose $A$ and $B$ are directed sets, and $\{A_b\}_{b \in B}$ is a collection of residual subsets of A indexed by B. Does there necessarily exist a monotone choice function $f: B \to A$ such that $f(b) \in A_b$ for each $b$? If so, can that function be made cofinal?

Edit: Thanks Stefan for showing the answer is no in general (I'll mark it as answered in a bit), but a follow up question: What if $b \ge b'$ implies $A_b \subset A_{b'}$? Is there still a counterexample in this case?

The background for this question comes from topology. Suppose $g:X \to Y$ is an open map, $g(x) = y$, and $\{y_a\}_{a \in A}$ is a net converging to $y$. Let $B$ index the open subsets of $X$ containing $x$, ordered by reverse inclusion. Every open neighborhood $U_b \ni x$ maps forward to a neighborhood of $y$, which contains all $y_a$ for $a$ in a residual set $A_b \subset A$. Can I create a net $\{x_b\}$ with $x_b \in U_b$ such that $g(x_b) = y_{f(a)}$? Can this be done such that the image is a subnet of $\{y_a\}$?

Answer 1

There is an easy counterexample which might help to guide ones mind to further examine when such monotone choice function can exist:

Let $B = (- \mathbb N; \le)$, $A = (\mathbb N; \le)$ and, for all $n \in \mathbb N$, $$ \tag{$\dagger$} A_{-n} = \mathbb N \setminus \{0,1, \ldots, n \}.$$

Any monotone choice function $f \colon B \to A$ for $(A_{n} \mid n \in - \mathbb N)$ induces an infinite decreasing sub-sequence in $(\mathbb N; \le)$. Hence there is no monotone choice function.

$(\dagger)$ It doesn't really matter what $A_{-n}$ is - we only need that those sets eventually disagree so that $f$ cannot be constant on an infinite set.

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OP asked in a comment:

Follow up: what if $b≥b′$ implies $A_b \subset A_{b'}$? Can you still find an example in that case?

With this extra assumption, there is always a monotone choice function: Fix, for each $b \in B$, some $a_b$ such that $$ \{ a \in A \mid a_b \le_A a \} \subseteq A_b. $$ Since $(A; \le)$ is directed, the additional assumption allows us to pick $(a_b \mid b \in B)$ such that $b \le_B b' \implies a_b \le_A a_{b'}$. Then $$ f \colon (B; \le_B) \to (A; \le_A), b \mapsto a_b $$ is as desired.