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This question has already been asked over here.

I was following this answer, but I'm having difficulties to prove that (with the notation given in the linked answer) for the maximal element $\langle M,\preceq_M\rangle$ given by Zorn's Lemma, we have $M=P$.

I tried to do it by contradiction. If $M\subsetneq P$, there must be some $m\in P-M$. If no element of $P-\{m\}$ is comparable to $m$ with respect to $\leq$, then one can show that $\langle M,\preceq_M\rangle$ isn't maximal (just take the sum $\langle N,\preceq_N\rangle$ of $\langle M,\preceq_M\rangle$ and $\langle \{m\},\{(m,m)\}\rangle$, for which one has $\langle M,\preceq_M\rangle\sqsubseteq \langle N,\preceq_N\rangle$).

How can one proceed if some element of $P-\{m\}$ is comparable to $m$ with respect to $\leq$?

Zuy
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1 Answers1

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The trick is to insert $m$ into an appropriate Dedekind cut in $M$.

Let $L=\{x\in M:x<m\}$ and $R=\{x\in M:m<x\}$. For each $x\in L$ and $y\in R$ we have $x<m<y$, so $x<y$, and therefore $x\prec_M y$. For $x\in M$ let $x\!\!\downarrow=\{y\in M:y\preceq_M x\}$, and let $L^+=\{m\}\cup\bigcup_{x\in L}x\!\!\downarrow$. Let $M^+=M\cup\{m\}$ and

$$\preceq_{M^+}=\preceq_M\cup\{\langle x,m\rangle:x\in L^+\}\cup\{\langle m,x\rangle:x\in M\setminus L^+\}\,;$$

then $\langle M^+,\preceq_{M^+}\rangle\in\mathscr{L}$, and $\langle M,\preceq_M\rangle\sqsubset\langle M^+,\preceq_{M^+}\rangle$, contradicting the maximality of $\langle M,\preceq_M\rangle$.

Brian M. Scott
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