Does there exist a full and faithful embedding of the category of posets into the category of sets? I suspect no, but I don't know how to prove or disprove this.

27$\begingroup$ The poset $\{0,1\}$ with $0<1$ has three endomorphisms. There is no set with exactly three endomorphisms. $\endgroup$ – Julian Rosen Dec 21 '18 at 3:03

4$\begingroup$ @JulianRosen you should add this as an answer, it's neat :) $\endgroup$ – David Roberts Dec 21 '18 at 4:13

4$\begingroup$ Julian’s answer is better, but also: there are infinitely many isomorphism classes of posets with a unique automorphism, but only two such sets. $\endgroup$ – Dylan Wilson Dec 21 '18 at 7:00

5$\begingroup$ A philosophy is that any full and faithful embedding $C \to D$ realizes the objects of $C$ as special types of $D$objects ($D$objects satisfying a certain property); see ncatlab.org/nlab/show/stuff,+structure,+property. Thus posets would have to be certain types of sets. This leads to considerations like the nice answers by Julian and Dylan. $\endgroup$ – Todd Trimble♦ Dec 21 '18 at 8:36

$\begingroup$ I thought @JulianRosen's answer was neat, too! A more philosophical question: why do so many people still seem to think that all mathematical objects are sets? The category of topological spaces with morphisms as homotopy classes of continuous functions is not even concretizable. $\endgroup$ – Jesse Elliott Dec 23 '18 at 0:48
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The poset $\{0,1\}$ with $0<1$ has three endomorphisms. Since there is no set with exactly three endomorphisms, there cannot be a fully faithful embedding of posets into sets.