# In a partial order does every finite subset have a maximum?

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.

• 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. – Nicolas Feb 15 at 9:09
• If you abandon the total order requirement you are abandoning a very crucial feature to allow for greatest elements to exist. – Git Gud Feb 15 at 9:11
• @GitGud total order=partial order+every pair of element is comparable(Right?) – StammeringMathematician Feb 15 at 9:12
• I dislike the way you write, but the idea is correct, that's exactly its definition. – Git Gud Feb 15 at 9:14

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.