I have the following problem:

Let $(A,\leq)$ a poset. Prove that there exist a total order $\leq^ *$ on $A$ such that if we have $a\leq b$ we can conclude $a\leq^*b$. (Hint: Use the Zorn Lemma)

I tried to use the hint. Considering the set $D$ of all the pair $(B,\leq _B)$ such that $B\subset A$ and $\leq_B$ is a total order such that $a\leq b$ implies $a\leq_B b$. I defined a partial order on $D$ in the following way: $$(B,\leq_B)\leq (C, \leq_C)$$ iff $B\subset C$ and $\leq_B\subset \leq_C$. I can prove that the hypothesis of the lemma Zorn holds. Then I can suposse that there exist maximal $(\bar A, \leq_{\bar A})$ in $D$. How do I prove that $A=\bar A$?

If I suposse that $A\not = \bar A$ then I can choose a $x\in A-\bar A$. I need specifically a way to construct a total order $\bar\leq$ on $A\cup \{x\}$ such that $\leq_{\bar A}\subset\bar\leq$ and if $a\leq b$ where $a,b \in A\cup \{x\}$ then $a\bar\leq b$. How can I do that?


  • 1
    $\begingroup$ possible duplicate of Every partial order can be extended to a linear ordering $\endgroup$
    – Git Gud
    Aug 26, 2014 at 11:07
  • 2
    $\begingroup$ THe problem is the same but I am asking not for the problem itself. $\endgroup$
    – EQJ
    Aug 26, 2014 at 11:09
  • $\begingroup$ @YTS Did you eventually find how to show that $\bar{A}=A$? $\endgroup$ Jun 28, 2018 at 9:30

1 Answer 1


You have found a maximal element $(\bar A, \leq_{\bar A})$ of your partial order on the pairs $(B, \leq_B)$ and you now want to prove that $\bar A = A$.

For a given $a \in A$, look at the trivial total order $\{(a,a)\}$ on $\{a\}$. Then $$ (\{a\}, \{(a,a)\}) \leq (\bar A, \leq_{\bar A}). $$ So, in particular, $a \in {\bar A}$.

  • $\begingroup$ I don't believe that this is correct. It assumes that $(\bar{A}, \leq_{\bar{A}})$ is the greatest element of $\leq$, or at least an upper bound of $D$. Rather, it is a maximal element, which is not the same thing. Unless you can show otherwise, it could very well be that $(\{a\}, \{(a,a)\})$ and $(\bar{A}, \leq_{\bar{A}})$ are incomparable in the order $\leq$ since it's not known to be a total order. $\endgroup$
    – kyp4
    Dec 16, 2017 at 19:31

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