For some function $f : \mathbb{N} \to S$, is there a concise (and preferably established) way to build tuples from sets $A\subset \mathbb{N}$ of the values they map to in $S$? The exact order of the tuple doesn't really matter, as long as it preserves some ordering.
An example is in place. Let $A = \{1, 5, 6\}$, I'm looking for a way to express the $|A|$-tuple $(f(1), f(5), f(6))$. The tuple $(f(5), f(6), f(1))$ is equally acceptable, but the (multi)set $\{f(1), f(5), f(6)\}$ would not do.
If we see tuples as nested ordered pairs $(a_1, a_2, \cdots, a_n) \equiv (a_1, (a_2, \cdots, a_n))$ with $(a_i)\equiv (a_i, \emptyset)$, I'm essentially looking for notation for the function:
$$ g(f, A) = \left\{\begin{array}{ll}
\emptyset & \text{if } A = \emptyset, \\
(f(\min A), g(A \setminus \min A)) & \text{else}.
\end{array}\right.$$
I'm tempted to use notation parallel to the set-builder, i.e., $(f(x) : x \in A)$ but that doesn't really convey that the ordering matter.
Edit to clarify the order requirement: The order of the tuple doesn't matter at all, but it matters that it is ordered. In other words, for $f : \mathbb{N} \to S$ and $h : \mathbb{N} \to S$, I want the following property of the notation:
$$(f(x) : x \in A) = (h(x) : x \in A) \iff \forall x \in A, f(x) = h(x).$$
Obviously, this property holds for any ordering as long as it only depends on $A$. But it doesn't really feel that $(f(x) : x \in A)$ conveys that.
Best Answer
Given $A \subseteq \mathbb{N}$, define $\ell_n(A) \in \mathbb{N}$ for $n < |A|$ inductively by $$\ell_0(A) = \mathrm{min}(A) \quad \text{and} \quad \ell_n(A) = \mathrm{min} \Big(A \setminus \{ \ell_0(A), \dots, \ell_{n-1}(A) \}\Big) \text{ for all } 0 < n < a$$ Thus the sequence $(\ell_n(A) \mid n < |A|)$ lists the elements of $A$ in increasing order.
Given a function $f : \mathbb{N} \to S$ and a subset $A \subseteq \mathbb{N}$, the sequence you seek is therefore $$\big( f(\ell_n(A)) \mid n < |A| \big)$$
Note that this definition is valid even when $A$ is empty, in which case it produces the empty sequence, and when $A$ is infinite, in which case it produces an infinite sequence of elements of $S$.