Notation for the set of all tuples whose elements are all part of a particular set

Question

So, for example, let's say we have A = {3,6,1} I would like to generate a set B which is such that the elements of B are tuples. Each tuple is such that if x is an element of the tuple then x is an element of A. Additionally, there are no repeated elements in any given tuple (each element of a given tuple is distinct). If A contains x number of elements, then B should contain all the tuples (meeting the distinctness restriction) with 1 element, all the tuples with 2 elements, ...., all the tuples with x elements.

So in our example, B would be: B = { (3) , (6) , (1) , (3,6) , (3,1) , (6,3) , (6,1) , (1,6), (1,3) , (3,6,1) , (3,1,6) , (6,3,1) , (6,1,3) , (1,3,6) , (1,6,3) }

so basically if A has x amount of elements, B is like the set of all permutations of A, taken x elements at a time, taken x-1 elements at a time, taken x-2 elements at a time,..., taken 1 element at a time. ....but instead of basic sets we have tuples.

Now, furthermore, I would like to have another set, C, which is such that x is an element of C iff x is an element of B, and x is such that its first element is less than its second (if there is a second), its second element is less than its third (if there is a third), etc...

(with "less than" being the standard less than relation among integers)

So we would get: C = { (3) , (6) , (1) , (3,6) , (1,6) , (1,3) , (1,3,6) }

Thanks

Answer 1

Interpreting the $n$-tuples with no repeated entries as injective functions from $[n]=\{0,1,2,\dots,n-1\}$ to $A$, and adopting the notation that $Y^{\underline{X}}$ is the set of all injective functions from $X$ to $Y$, then you could notate it as:

$B = \bigcup\limits_{n=1}^\infty A^{\underline{[n]}}$

As for denoting $C$, I think you would have an easier time referring instead to the power set of $A$ (without the emptyset) and reinterpret each set appearing as a tuple with entries in ascending order.

$C\simeq \mathcal{P}(A)\setminus\{\emptyset\}$

In terms of counting how large each set is, $|B|=\sum\limits_{n=1}^\infty |A|^{\underline{n}}$ using falling factorial notation and $|C|=2^{|A|}-1$ (if you removed the emptyset)