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I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for every partial order (P, ≤) ∈ V with the property that there exists a well- order ≺ on the underlying set P, there is an inclusion-maximal chain X in (P, ≤). How could I prove it using only ZF and not Choice?

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    $\begingroup$ Do you happen to have heard that in problem set 5? $\endgroup$ – Alessandro Codenotti Nov 14 '18 at 9:18
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Recursively construct $(C_\xi \mid \xi \in \mathrm{Ord})$ as follows:

Let $C_0 := \emptyset$ and given $C_\xi$ either $C_\xi$ is a $\subseteq$-maximal chain in $(P; \le)$ in which case we stop the construction or otherwise $$ C_{\xi +1} := C_\xi \cup \min_{\prec} \{ p \in P \setminus C_\xi \mid \forall c \in C_\xi \colon p \le c \vee c \le p \}. $$ For limit $\lambda \in \mathrm{Ord}$, we let $C_{\lambda} := \bigcup_{\xi < \lambda} C_{\xi}$.

It is easy to verify that each $C_\xi$, if defined, is a chain through $(P; \le)$ and that there is some $\xi < H(P)$ such that $C_{\xi}$ is a maximal chain.

(Here $H(P)$ is the least ordinal $\alpha$ such that there is no injection $i \colon \alpha \to P$. Since $P$ has a well-ordered, we have that $H(P) = \mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)

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  • $\begingroup$ What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all. $\endgroup$ – Asaf Karagila Nov 13 '18 at 20:24
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    $\begingroup$ @AsafKaragila As you can see, I've mentioned that. $\endgroup$ – Stefan Mesken Nov 13 '18 at 20:25
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    $\begingroup$ As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P $\endgroup$ – Asaf Karagila Nov 13 '18 at 20:26

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