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 '14 at 11:07
  • 2
    $\begingroup$ THe problem is the same but I am asking not for the problem itself. $\endgroup$ – YTS Aug 26 '14 at 11:09
  • $\begingroup$ @YTS Did you eventually find how to show that $\bar{A}=A$? $\endgroup$ – Daniele1234 Jun 28 '18 at 9:30

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 '17 at 19:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.