# A question about proof by contradiction

$1$. Claim: $\varnothing$ is an antichain.

$Proof$: Suppose $\varnothing$ is not an antichain. Then $\exists$ a pair $x, y \in \varnothing$ such that $x$ and $y$ are comparable. Contradiction: $\varnothing$ is empty. Thus $\varnothing$ is an antichain.

$2$. Claim: $\varnothing$ is a chain.

$Proof$: Suppose $\varnothing$ is not a chain. Then $\exists$ a pair $x, y \in \varnothing$ such that $x$ and $y$ are incomparable. Contradiction: $\varnothing$ is empty. Thus $\varnothing$ is a chain.

What am I doing wrong here? Or is $\varnothing$ both a chain and antichain?

• You’re doing nothing wrong: $\varnothing$ is vacuously both a chain and an antichain. Commented Nov 19, 2014 at 23:20
• See here Commented Nov 19, 2014 at 23:22
• The empty set is an ordered set which is both a chain and an antichain. Nothing contradictory in this. In fact the same can be said about a singleton $\{ x \}$ with the trivial order. Commented Nov 19, 2014 at 23:25
• Thanks, y'all. I am just trying to make sure I am not making up stuff. Commented Nov 19, 2014 at 23:31

The logical error is the conclusion of the two proofs. Your specific definitons of a chain and an antichain together do not cover all the possibilities, so "is not a chain" does not necessarily imply "is an antichain." Using the specific definitions you've given would imply that $\emptyset$ is neither a chain nor an antichain.