In Lipschutz, Theory and Problems of Set Theory ( Schaum's series, 1964 ed.), Chapter 10, Solved Problem n°7 , the question is asked to find all non-empty totally ordered subsets of an ordered set A = {a,b,c} with a diagram showing that : b precedes a and that c precedes a ( b and c being incomparable).

Lipschutz provides a solution in which he gives {a} , {b} and {c} as totally ordered subsets of A.

Hence my question : is every singleton subset of a partially ordered set a totally ordered set.

How to explain that , in case a subset is a singleton the trichotomy condition for totally ordered sets is automatically fullfilled : a partial order S is total iff for all a and b belonging to S,

                      either a<b or a=b or a>b. 

I'd like to derive the conclusion formally from the definition of a totally ordered set.

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    $\begingroup$ It's because $\forall a,b: a=b$ holds in a singleton. $\endgroup$ – Berci Apr 29 at 12:27
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    $\begingroup$ $a \in \{ a \}$ is the only element and $a=a$. Thus, the trichotomy condition is fulfilled. Recall that the condition id $\forall a,b \in X \ [a=b \lor \ldots ]$. $\endgroup$ – Mauro ALLEGRANZA Apr 29 at 12:28
  1. Let $S$ be a singleton.
  2. From $1$, there follows there must be some $x$ such that $S=\{x\}$.
  3. Let $a, b\in S$.
  4. Therefore, $a=x$ and $b=x$.
  5. Therefore, $a=b$.
  6. Therefore, $a=b\lor a<b\lor a>b$.
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    $\begingroup$ Very frustrating if nice answers like this are downvoted. And this without any explanation why. $\endgroup$ – drhab Apr 30 at 8:34
  • $\begingroup$ @dmab. Totally agree. I personnaly upvoted both aswers. $\endgroup$ – Eleonore Saint James Apr 30 at 9:25

Yes, every singleton can be recognized as a totally ordered set (even as a well-ordered set).

This because next to reflexivity, transitivity and antisymmetry we also have comparability.

If $S=\{c\}$ is a singleton then it is true that for all elements $a,b\in S$ we have: $$a<b\text{ or }a=b\text{ or }a>b$$

Actually in all cases (there is only one) we have $a=c=b$.

Also note that every non-empty subset of $S$ (there is only one) has a smallest element.


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