Definition: A partition of a set $A$ is a set of nonempty subsets of $A$ called the blocks of the partition, such that every element of $A$ is in some block, and if $B$ and $B'$ are different blocks, then $B \cap B' = \emptyset$
Let $\ R \ $ be a strict partial order (irreflexive) relation on set $A$,
$depth(a)$: length of the longest chain ending at $a$, where $a \in A$
$A_k$ ::= $\{a| \ depth(a) = k \}$
Proposition: I want to show that all possible sets $A_k$ for $k \in \mathbb{N}$, form a partition of set $A \ $,
i.e. I want to show that every element, $a$, of set $A$ is a member of some block $A_k$, and the set $A_k$ has no intersection with other set $A_j$ where $j \neq k$
I have the following concerns:
If $A$ is a finite set, then definitely length of the longest chain that ends at $a$ can not be greater than $|A|$. So always $a \in A_k$ for some finite $k \in \mathbb{N}$
But I have concerns when the set $A$ is countably infinite or uncountable. I tried to apply principle of infinite descent for the countable case by arguing that there exists a bijection between set $A$ and naturals but could not succeed.
And I do not think Proposition is valid if $A$ is uncountable.
Help needed for proof and validity of proposition for both countable and uncountable cases of set $A$.
As far as your definitions make sense for infinite sets, the proposition is false. For instance, in $\mathbb Z$ with the strict partial order $\lt$, there are infinite chains ending at every element. To make sense of your definition, we'd have to say that the depth of every element is infinite. But then none of the elements are in any of the $A_k$ for $k\in\mathbb N$.