Large ordered family of set inclusions complete?

Question

I'm not an expert on set theory, so this might be trivial: Let $\dots \subseteq X_{\alpha} \subseteq X_{\alpha+1} \subseteq \dots$ be a chain of set inclusions, indexed by ordinals(!), s.t. $X_{\alpha}\subseteq X$ for all $\alpha$. Does there exist a supremum of this chain?

Answer 1

Your set sequence is eventually constant: we can find some ordinal $\gamma$ such that for all $\alpha>\gamma$, $A_\gamma = A_\alpha$.

Its easiest proof goes as follows: Assume that the sequence is not eventually constant. Without loss of generality, we can assume the sequence is strictly increasing, by removing sets identical to the previous ones in the sequence. Fix a well-ordering $\prec$ over $A$ and take $a_\xi \in A_{\xi+1}-A_\xi$ be the $\prec$-minimum of $A_{\xi+1}-A_\xi$. We can see that, if $\xi\neq \eta$ then $a_\xi\neq a_\eta$ so $A$ contains a set equipotent to the class of all ordinals, which is impossible.

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Above proof uses the well-ordering principle, but we can avoid it. Assume to the contrary, that our sequence is not eventually constant. We can also assume that the sequence is strictly increasing. Let define $f$ from $A$ to the class of ordinals by stipulating $$f(x) = \begin{cases}\alpha&\text{if }x\in A_{\alpha+1}-A_\alpha \\ 0 &\text{otherwise}\end{cases}.$$ Then $f$ is a surjection from $A$ to $\mathrm{Ord}$. However, we can prove that it is impossible. Mimicking the proof of the Hartogs theorem provides the nonexistence of such $f$.