I wonder if I have taken the correct approach to demonstrating what ∪∪7 is, using Von Neumann's proposal? And if I am indeed correct? My answer appears a little odd to me!
Question: What is ∪∪7?
By the successor function: r+ = r ∪ {r} ⇒ ∅+ = ∅ ∪ {∅} ⇒ ∅+++++++ = ∅ ∪ {7}
∪7 = ∅, {∅} ⇒ ∪∪7 = ∅
Thankyou for your help!
During this answer, when I say "ordinal," I will at all times mean "von Neumann ordinal," i.e.: a transitive set that is well-ordered by $\in$.
Lemma: If $X$ is a set of ordinals, then $\bigcup X$ is the smallest ordinal $\gamma$ such that $\gamma\geq\alpha$ for all $\alpha\in X$. In particular if $X$ has a greatest element, then it is $\bigcup X.$
Proof: Since each ordinal is a set of ordinals, then $\bigcup X$ is a class of ordinals, and so is well-ordered by $\in$. Furthermore, since $X$ is a set, then $\bigcup X$ is a set by Axiom of union. Also, if $\alpha\in \bigcup X,$ then $\alpha\in\beta$ for some $\beta\in X,$ so since $\alpha\subseteq\beta$ and $\beta\subseteq\bigcup X,$ then $\alpha\subseteq\bigcup X,$ and so $\bigcup X$ is transitive. Thus, $\bigcup X$ is an ordinal (say $\gamma$), since it is a transitive set well-ordered by $\in$.
Since $\alpha\subseteq\bigcup X=\gamma$ whenever $\alpha\in X,$ then $\alpha\leq \gamma$ whenever $\alpha\in X.$ Now, suppose that $\beta$ an ordinal such that $\beta\geq\alpha$ whenever $\alpha\in X$. Then $\alpha\subseteq\beta$ whenever $\alpha\in X,$ so $\gamma=\bigcup X\subseteq\beta$, and so $\gamma\leq\beta,$ as desired. $\Box$
You seem to be a bit confused about the successor function. The first few:
$$0:=\emptyset\\1:=0^+=0\cup\{0\}=\{0\}\\2:=1^+=1\cup\{1\}=\{0,1\}\\3:=2^+=2\cup\{2\}=\{0,1,2\}\\4:=3^+=3\cup\{3\}=\{0,1,2,3\}\\\vdots$$
More generally, any ordinal $\alpha$ is the set of all of its predecessors, and in particular, every non-$0$ finite ordinal $n$ has a greatest element, which (by the Lemma) is $\bigcup n$.
In particular, $\bigcup7=6,$ and $\bigcup\bigcup7=\bigcup6=5.$