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An ordered set can be regarded as a category. What are its subcategories? Which of them are full?

Here's what I understand.

The elements of the ordered set correspond to the objects of the category. There are at most one arrow between any two objects, and there is an arrow $a\to b$ iff $a\le b$ in the ordered set. A subcategory consists of a subset of object and a subset of arrows subject to some conditions. So at least a subcategory is a subset of the ordered set. It also comes with a collection of arrows. For any two objects $a,b$ in the subcategory, if there is no arrow from $a$ to $b$ in the original category, then there cannot be arrow in the subcategory. If there is an arrow $a\to b$ in the original category, then this arrow may or may not belong to the subcategory. In particular, it cannot be the case that there is an arrow $a\to b$ in the subcategory but there is no arrow $a\to b$ in the original category. If $a\to b$ and $b\to c$ are in the subcategory, then $a\to c$ must be in the subcategory. But I don't see how to extract a reasonable classification of subcategories (including full subcategories) from all of the above.

There is a question that includes my question: What are the subcategories of ordered sets / groups?

But I don't understand the answer (see my comment to the accepted answer and see the above for an explanation/proof of the claim of the comment).

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You're just about there. Your subcategory consists of a bunch of objects with at most one arrow between any two of them. That makes it a candidate for being an ordered set. Now check whether the arrows in your subcategory must fulfill the conditions to actually be an ordered set.....

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  • $\begingroup$ If you are talking about an arbitrary subset of objects and arrows of the original category such that the composition of two composable arrows are in this subset, then the properties of reflexivity and transitivity automatically hold. If we want this ordered subset to satisfy antisymmetry, then it means that for no two distinct objects $a,b$ can one find arrows $a\to b$ and $b\to a$. But I don't see how this gives the desired classification. $\endgroup$ – user634426 Jun 25 '19 at 1:53
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This is a little bit expanded version of this answer; thanks to Giorgio Mossa for the discussion in the comments.

Let $\mathscr P$ be a category corresponding to a poset $(P,\le_P)$. Consider a subset $Q\subseteq P$ of the set of objects of $\mathscr P$. Assume it is the set of objects of a subcategory $\mathscr Q$ of $\mathscr P$. Since there are at most one arrow between any two objects of $\mathscr P$, there is also at most one arrow (in $\mathscr Q$) between any two elements of the subset $Q$ we are considering. Define the relation $\le_Q$ on the set $Q$ as follows: $a\le_Q b$ iff there is an arrow from $a$ to $b$ in $\mathscr Q$. We claim that $\le_Q$ is a partial order.

Since $\mathscr Q$ contains all identity arrows for every object of $\mathscr Q$, the relation $\le_Q$ is reflexive. It is also transitive since $\mathscr Q$ is closed under composition. The anti-symmetry of $\le_Q$ follows because if $a\to b$ and $b\to a$ in $\mathscr Q$ with $a\ne b$, then we also have $a\to b$ and $b\to a$ in $\mathscr P$ with $a\ne b$, which contradicts the fact that $\le_P$ is anti-symmetric. These three facts show that $\le_Q$ is a partial order.

Now note that if there were no arrow $a\to b$ in $\mathscr P$, then there would not be such an arrow in $\mathscr Q$ either. Taking the contrapositive and translating into the language of orders, this means that if $a\le_Q b$, then $a\le_P b$.

Therefore, the subcategories of $\mathscr P$ are the categories corresponding to a poset $(Q,\le_Q)$ with the property that for all $a,b\in Q$, one has $a\le_Q b\implies a\le_P b$, together with the empty subcategory.

A subcategory $\mathscr Q$ of $\mathscr P$ is full if for all $a,b\in Q$, if $a\to b$ in $\mathscr P$, then also $a\to b$ in $\mathscr Q$. In the language of orders, this means that for all $a,b\in Q$, if $a\le_P b$, then $a\le_Q b$. Thus the full subcategories are those corresponding to a poset $(Q,\le_Q)$ with the property that for all $a,b\in Q$, one has $a\le_Q b\iff a\le_P b$, together with the empty subcategory.

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