I have solved the following problem without using a given premise. Could someone please spot whether I have done something wrong?
Suppose we have a relation $\geq$ that is transitive, but not necessarily complete. For nonempty $A$, we define $$ C(A)=\{x\in A: x\geq y\text{ for all }y\in A\},\\ D(A)=\{x\in A: \text{there exists no }y\in A\text{ such that }y>x\}. $$ Prove that $C(A)\subset D(A)$. Prove that $\geq$ is complete iff $D(A)$ is nonempty for all finite $A$.
Proof. For the first claim, suppose $x\in C(A)$. If $\exists y\in A$ such that $y>x$ then, in particular, $\neg(x\geq y)$, contradicting $x\in C(A)$. This means $x\in D(A)$.
The second claim is where I might have made a mistake: I claim that $D(A)$ is always nonempty as long as $A$ is finite; that is, completeness is irrelevant. Suppose otherwise that $D(A)$ is empty and consider $x_1\in A$. Then, there exists $x_2\in A$ such that $x_2>x_1$. In turn, there must be $x_3\in A$ such that $x_3>x_2$. In this manner, we can build a sequence $\{x_n\}_{n=1}^\infty$ in which $x_{n+1}>x_n$ and transitivity implies that the elements of the sequence are distinct. But this violates the finiteness of $A$, giving us our contradiction. ∎
Appendix. Some definitions for those unfamiliar with this context:
- Transitivity: $x\geq y$ and $y\geq z$ imply $x\geq z$.
- Completeness: $\forall x,y$, either $x\geq y$ or $y\geq x$ (or both).
- $x>y$: $x\geq y$ and $\neg(y\geq x)$.
Your reasoning is impeccable. Looks like a typo in the problem. The last line should read: Prove that $\ge$ is complete iff $C(A)$ is nonempty for all (nonempty) finite $A$.