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I know that for any two sets $A,B$, it holds that $A\subseteq B$ iff every element of $A$ is in $B$, intuitively. I also know that it is reflexive, antisymmetric and transitive.

But, is it technically a poset? In the wiki page of a poset says that the subset relation is defined on the power set of a set, but for that we need some universe set $U$, and then define the subset relation. But clearly, subset relation is defined for all sets, which is a proper class, not a set, and because there is no set of all sets, we cannot define a universe set, so I feel this definition is ill-formed.

So, what is the correct way of defining a subset "relation", If not a poset on the class of sets?

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  • $\begingroup$ I would use the term quasi-poset, but that is personal. I never met that term. $\endgroup$
    – drhab
    Commented Mar 20, 2020 at 11:00

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The answer is that you would restrict this to some universe that is a set if you want to get a poset. Otherwise the relation, while it exists, is a proper class and hence does not define a partially ordered set. In most subjects we are only concerned with sets contained in some larger set, hence the collection of all of the ones we care about forms a set. Otherwise we still have a relation that makes sense, and we can still analyze it, but as you said it does not define a poset.

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  • $\begingroup$ What if I want to talk about the subset relation in the category $\textbf{Rel}$? Am I allowed to talk about the subset relation as a(transitive,antisymmetric, reflexive) morphism(relation) between sets? $\endgroup$
    – Garmekain
    Commented Mar 20, 2020 at 11:01
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    $\begingroup$ @Garmekain The collection of relations between two sets $A$ and $B$ is always a subset of the power set of $A\times B$. $\endgroup$ Commented Mar 20, 2020 at 11:09
  • $\begingroup$ Oh I was incorrect then. Thanks. $\endgroup$
    – Garmekain
    Commented Mar 20, 2020 at 11:10
  • $\begingroup$ @Garmekain No problem. $\endgroup$ Commented Mar 20, 2020 at 11:11
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You can still define order relations on a class that is not a set, but it is then not a partially ordered set.

See: https://en.wikipedia.org/wiki/Preordered_class

Given a set $X$, it is true that $(\mathcal{P}(X), \subseteq)$ is a partially ordered set.

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