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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