#proofinatoot that if a finite poset has a unique maximal \(x\), then \(x\) is maximum.
If not, there is a \(y_1||x\). \(y_1\) is not maximal, so there is \(y_2>y_1\); we cannot have \(y_2<x\), else transitivity would give \(y_1<x\), and we cannot have \(y_2>x\) because \(x\) is maximal, so \(y_2||x\). Continuing, we build a chain \(y_1<y_2<\dotsb\) (with \(y_i||x\) for all \(i\)), contradicting finiteness.
(This proof also suggests a construction of an infinite poset without the property.)
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