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If $≤$ is a total ordering on A, then every non-empty finite subset S of A has a least element and a greatest element.

I was wondering whether this result is true if we replace "total ordering" by partial ordering.

Do we have some example. Thanks for help and reading out.

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  • $\begingroup$ Hint: Use induction over $|S|$ to show that $S$ has a greatest (or a least) element. For a counterexample, see drhab's answer or consider any partial order where there is more than one minimal or maximal element. $\endgroup$ – Nicolas Feb 15 at 9:09
  • $\begingroup$ If you abandon the total order requirement you are abandoning a very crucial feature to allow for greatest elements to exist. $\endgroup$ – Git Gud Feb 15 at 9:11
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    $\begingroup$ @GitGud total order=partial order+every pair of element is comparable(Right?) $\endgroup$ – StammeringMathematician Feb 15 at 9:12
  • $\begingroup$ I dislike the way you write, but the idea is correct, that's exactly its definition. $\endgroup$ – Git Gud Feb 15 at 9:14
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Not true in general for partial order.

Counterexample: set $A$ with more than one element and equipped with partial order $=$.

If $a,b\in A$ with $a\neq b$ then the set $\{a,b\}$ has no least and no greatest element.

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