Is the sequence of sets $\cup X, \cup \cup X, \cup^3 X, \cup^4 X, \cdots$ eventually constant for all sets $X$? Is the sequence of sets $\cup X, \cup \cup X, \cup^3 X, \cup^4 X, \cdots$ eventually constant for all sets $X$?
I'm assuming that we're working with ZFC.
First, I am going to nonstandardly insist that  $\cap \varnothing$ is equal to $\varnothing$. I am only doing this so that $\cap$ is total when thought of as operation from set to set. In addition, $\circ$ is composition of relations.
We know by the axiom of foundation that the following is true,
$ A_1, A_2, A_3, A_4, A_5, \cdots $, where $A_{i+1} \in A_{1}$ is finite.
If we ever hit $\varnothing$ or an infinite cardinal, then we have hit a fixed point. I'm not sure whether these are all the fixed points of $\cup$, however, or if we are guaranteed to hit such a fixed point if we start with an arbitrary set.
If my set is hereditarily finite, then repeatedly unioning it will decrease the length of the longest descending chain by one each time (I think?) ... so we'll eventually end up at $\varnothing$.
If we take the following sequence $B$ for any set $X$,
$$ \cap X, {\cap}{\cap} X, \cap^3 X, \cap^4 X, \cap^5 X\cdots $$
We know that, for all $i$, either $B_{i+1} \subset \circ \in B_i $ or $B_i = B_{i+1} = \varnothing$.
The relation $\subset \circ \in$ is a composition of a linear order and a well-order. Therefore $\subset \circ \in$ is a well-order. Therefore $B$ is eventually constant, and the value that it eventually settles on is $\varnothing$.
I'm curious about the sequence $C$ below.
$$ \cup X, {\cup} {\cup} X, \cup^3 X, \cup^4 X \cdots $$
$\cup$ has some fixed points besides $\varnothing$. In particular, the ordinal $\omega$ is the set of all finite ordinals, and the following is true.
$$ \omega = \cup \omega $$
We can prove this by noticing that every finite ordinal is an element of its successor and every finite ordinal contains exclusively finite ordinals as its elements.
So this means that $C$ is eventually constant when $X$ is $\omega$.
I think $C$ is eventually constant whenever we are starting with an infinite cardinal or a hereditarily finite set. For the later case, the length of the longest chain decreases by one each time we union all the elements together ... I think.
Now I'm wondering whether $C$ is eventually constant regardless of my starting set.
 A: Define the Zermelo naturals recursively as $0=\emptyset,$ $n+1=\{n\}.$ Then let $X = \{1,2,4,8,16,32,\ldots\}$ be the powers of two. Then $\cup X = \{0,1,3,7,15,31,\ldots\},$ $\cup\cup X = \{0,2,6,14,30,\ldots\}$ and so on. This will not stabilize after a finite number of iterations since $X$ has arbitrarily large gaps between the numbers.
(Edit: Actually, now that I think about it, arbitrarily large gaps aren't necessary. For instance, the even numbers will work just as well.)
A: A simpler example has already been posted. For what it's worth, here's an example where
$$X\subsetneqq\bigcup X\subsetneqq\bigcup\bigcup X\subsetneqq\bigcup\bigcup\bigcup X\subsetneqq\cdots.$$
Define $f(A)=A$, $f_{n+1}(A)=\{f_n(A)\}$, $g_n(A)=\{f_n(A),f_{n+1}(A),f_{n+2}(A),\dots\}$.
Let $A_1,A_2,A_3,\dots$ be distinct infinite subsets of $\omega$, and let
$$X=\omega\cup g_1(A_1)\cup g_2(A_2)\cup g_3(A_3)\cup\cdots.$$
Then we have
$$\bigcup X=\omega\cup g_0(A_1)\cup g_1(A_2)\cup g_2(A_3)\cup\cdots,$$
$$\bigcup\bigcup X=\omega\cup g_0(A_1)\cup g_0(A_2)\cup g_1(A_3)\cup\cdots,$$
$$\bigcup\bigcup\bigcup X=\omega\cup g_0(A_1)\cup g_0(A_2)\cup g_0(A_3)\cup\cdots,$$
etc., so that $\bigcup^{n-1}X\subseteq\bigcup^nX$ and $A_n\in(\bigcup^nX)\setminus(\bigcup^{n-1}X)$.
