How can I prove that the transitive close of an arbitrary set $X$ exists?
A set $X$ is called transitive if $\in$ is transitiv on it. The transitive closure $\mathrm{tcl}(X)$ of an arbitrary set $X$ is the smallest transitive set (w.r.t. inclusion) which is a superset of $X$.
I guess it would suffice to show that any transitive superset of $X$ exists but I am failing right there.
Let me show you my try (note: I know that it fails). Define a function $G(n), n\in \Bbb N$ with
$$G(0)=X,\quad G(n+1)=\bigcup X.$$
This function exists because of the recursive function theorem. From there we can build the set $\mathcal G=\{G(n)\mid n\in \Bbb N\}$ by the axiom of replacement. We then have $\mathrm{tcl}(X)=\bigcup \mathcal G$. This works, because $\in$ is well-founded. Because if $\bigcup \mathcal G$ would not be transitive, this means ther would be an infinite descresing $\in$-sequence (not okay because of axiom of foudnation).
However, as I said, this fails.
- I cannot meaningully define $G$, because I do not realy have any clue what the codomain and domain could be. And I need them to define $G$ as a subset of $\mathrm{Cod}(G)\times \mathrm{Dom}(G)$.
- The part about why $\bigcup\mathcal G$ is transitive is a bit sloppy. But I might be able to make this rigorous. The main problem is the first item.
Best Answer
I like your question, because you found a subtle flaw in the way that people sometimes talk about this theorem.
Let's argue instead like this. We shall use the regularity axiom and argue by $\in$-induction. If there is a set without a transitive closure, then there is an $\in$-minimal such set. So we may assume without loss that every element $a\in X$ has a transitive closure $tc(a)$. By the replacement axiom, therefore, we may form the set $\{X\}\cup\{tc(a)\mid a\in X\}$. The union of this set is transitive and contains $X$ as a subset, and in fact, it is precisely the transitive closure of $X$.
An alternative argument would be to argue like this: $X$ is in some $V_\theta$ in the cumulative hierarchy, and every $V_\theta$ is transitive. So this provides a suitable co-domain for your funtion $G$.
Here is third way to argue. For each natural number $n$, there is a unique sequence of length $n$ that iterates the union operation starting from $X$. This can be proved by induction on $n$, since there cannot be least number $n$ violating this. Thus, by the axiom of replacement, we can take the set of all such sequences. This amounts to having your function $G$, from which one can construct the transitive closure.