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Let $\le$ a partial order on a set $A$. Prove that there exists a linear order $\le ^*$ on $A$ such that $a\le b$ implies $a\le ^*b$.

I considered $\mathcal{F}$ the family of all $\le _0 \subseteq A\times A$ such that $\le\subseteq\le _0$ and $\le _0$ is a partial order on $A$.

Then, using Zorn's lemma I concluded that $\mathcal{F}$ has a maximal element $\le ^*$. I proved that this is a partial order on $A$.

Now, I suppose this is not a linear order, then there exists $a,b\in A$ such that $(a,b)\not\in\le ^*$ and $(b,a)\not\in\le ^*$.

Where is the contradiction here? First, I considered $\le _0^* : = \le ^*\cup \{ (a,b)\}$ and I wanted to prove that this is a partial order, but I can't prove the transitivity.

Any hint? Thanks.

user74411
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  • Nearly a duplicate: http://math.stackexchange.com/questions/271003/every-partial-order-can-be-extended-to-a-linear-ordering/ – Asaf Karagila May 23 '13 at 17:12

3 Answers3

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HINT: You can add either $\langle a,b\rangle$ or $\langle b,a\rangle$, but you have to add more than the single ordered pair. To add $\langle a,b\rangle$, let $L(a)=\{x\in A:x\le^* a\}$ and $R(b)=\{x\in A:b\le^* x\}$, and show that

$$\le^*\cup\;\big(L(a)\times R(b)\big)\;$$

is a partial order.

Brian M. Scott
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Hint: You can't just arbitrarily include $(a,b)$. Set $M(a)=\{x:x\le a\}$, $M(b)=\{x:x\le b\}$, $N(a)=\{x:x\ge a\}$, $N(b)=\{x:x\ge b\}$. Look at these sets, and choose $(a,b)$ or $(b,a)$ based on them.

vadim123
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There is the theorem of Szpilrajn (see E.Harzheim, Ordered Sets, Springer, 2005):

Let $P(\le)$ be a poset. Then there exists a linear order $\le_*$ on $P$ which contains $\le$.

It is proved using the axiom of choice.

Boris Novikov
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