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There're are excellent answers by Asaf and Brian here. I have my own questions on this:

  1. Asking for an alternative proof approach: I am following Halmos's Naïve Set Theory, and he gives this as an exercise just after proving well ordering theorem, and not in the previous section dealing with Zorn's lemma. It's as if he intends to use well ordering theorem to prove this. And I am just scratching my head how well ordering might be used here to give a proof. Everywhere, Zorn's lemma seems to be used for this.

  2. Asking for a proof explanation: Brian's answer to the above linked question uses Zorn's lemma on the set $\mathscr L$, partially ordered by inclusion, of all the total orders extending the given partial order on the original set, say $P$. But the answer invokes Zorn's lemma by showing that any chain $\mathscr C$ in $\mathscr L$ has an upper bound in $\mathscr L$. The answer mentions to take the obvious choice for this upper bound, which I assume is $\bigcup \mathscr C$. But then this doesn't seem to work if $\mathscr C = \emptyset$. What am I missing? Note that what is to be proven is that $\mathscr L$ is nonempty.

drhab (in comments) has resolved the second question: Brian in his answer, means the restriction of $\le$ to $X\times X$, when he writes $\le\upharpoonright(X\times X)$. Hence $\mathscr L$ is the set of all the total orders on subsets $X$ of $P$ such that they extend the restriction of $\le$ on $X\times X$.

Atom
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  • If chain $\mathscr C=\varnothing$ then every element of $\mathscr{L}$ serves as an upper bound of the chain. So $\mathscr{L}\neq\varnothing$ is enough then. – drhab Jan 20 '22 at 12:58
  • But then, what we are proving is that $\mathscr L$ is nonempty. – Atom Jan 20 '22 at 13:00
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    Isn't it evident that $\mathscr L$ is not empty? You can take a singleton for $X$ equipped with equality relation. The aim is not to prove that $\mathscr L$ is not empty but that it has a maximal element. – drhab Jan 20 '22 at 13:07
  • @drhab But then, this does not contain the original order, does it? (Also, by $X$, you mean the given poset, right?) – Atom Jan 20 '22 at 13:57
  • The maximal element $M$ contains the original order $P$. This according to the third bullet in the answer of Brian. By $X$ I mean an element of $\mathscr L$ (defined in the answer of Brian). If $p$ is an element of $P$ then we can take $X=({p},=)\in\mathscr L$. – drhab Jan 20 '22 at 14:01
  • @drhab But $\mathscr L$ is defined to contain only those $\langle X, \preceq_X\rangle$ such that $\preceq_X$ is an extension (and not just compatible on the common elements of $P$ and $X$) of the order $\le$ on $P$, in the sense that $\preceq_X\supseteq\le$ (viewing these as subsets of $P\times P$). And most likely ${(p,p)}\nsupseteq \le$. Where am I not getting it? (Thanks for your patience!) – Atom Jan 20 '22 at 17:43
  • (See for example how Asaf defines a set in his answer.) – Atom Jan 20 '22 at 17:52
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    $X$ is a subset of $P$ equipped with a linear order $\preceq_X$. This order is not necessarily an extension of order $\le$ on $P$. It is an extension of $\le$ restricted to $X\times X$. If $X$ is a proper subset of $P$ then it is not even possible that we are dealing with an extension of $\le$ because in that case we do not have $\le\subseteq X\times X$. I am only talking about Brians answer here. – drhab Jan 20 '22 at 20:54
  • Again if $p\in P$ and $X={p}$ equipped with linear order $\preceq_X={(p,p)}$ then $\langle X,\preceq_X\rangle\in\mathscr L$. This because the restriction of $\le$ to $X\times X$ equals $\preceq_X={(p,p)}$ (so formally is extended by $\preceq_X$). – drhab Jan 20 '22 at 21:18
  • @drhab I would have had a sleepless night if it were not for you! – Atom Jan 20 '22 at 22:38

1 Answers1

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There's a recent proof by Michael Mandler, which I think is absolutely fantastic.

Mandler, Michael, A quick proof of the order-extension principle, Am. Math. Mon. 127, No. 9, 835 (2020). ZBL07264146.

Fix a partial ordered set $(X,\leq)$.

  1. $(X,\leq)$ embeds into $(\mathcal P(X),\subseteq)$ by $x\mapsto C_x=\{y\in X\mid x\leq y\}$.

  2. Well order $X$, so without loss of generality, $X=\alpha$ for some ordinal.

  3. $(\mathcal P(\alpha),\subseteq)$ has a natural linear order: $A\lhd B$ if and only if $A=B$ or $\min(A\mathbin{\triangle}B)\in B$. In other words, $A$ is smaller than $B$, if the first element not in both $A$ and $B$ lies in $B$.

  4. Note that $\lhd$ extends $\subseteq$, since $A\subseteq B$ if and only if $A\mathbin{\triangle}B\subseteq B$.

  5. Define $x\preceq y\iff C_x\lhd C_y$.

  6. Note that if $x\leq y\iff C_x\subseteq C_y$ and therefore $x\leq y\to x\preceq y$, so $\preceq$ is a linear extension of $\leq$.

Asaf Karagila
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