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One knows that the Alexandrov topology on a preordered set is the finest topology that induces the same [specialization] preorder on the set.

Given this, one finds a one-to-one correspondence between the Alexandrov topologies on a set and the pre-orders on that set.

On the other hand, every pre-order can be characterised by a thin category.

My question is:

If there's anyway to formulate the Alexandrov topology on the pre-ordered set totally algebraically in terms of the thin category?

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We just have to combine the two equivalences.

If $\mathcal{C}$ is a thin category, we call $U \subseteq \mathrm{Ob}(\mathcal{C})$ open if for every object $X \in U$ and any morphism $X \to Y$, we also have $Y \in U$. This declares an Alexandrov topology on $\mathrm{Ob}(\mathcal{C})$.

It is straight forward to check that this construction actually works for every small category. (Strange, I never have seen this topology before.)

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  • $\begingroup$ Thanks. You mean you've not faced this type of topology over a category in the mathematical literature? $\endgroup$
    – Bastam Tajik
    Commented Oct 6, 2023 at 12:04
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    $\begingroup$ Yes. But on the other hand, the construction also factors through the right adjoint $\mathbf{Cat} \to \mathbf{PreOrd}$. So nothing is gained from it. $\endgroup$
    – Martin Brandenburg
    Commented Oct 6, 2023 at 12:06
  • $\begingroup$ Do you think there's any relation between this topology and the nerve of this category? A nd in general do you mind if I improve the question and complexify it till we reach what I seek physically? I am a physics student. Can you improve your answer respectively too? $\endgroup$
    – Bastam Tajik
    Commented Oct 6, 2023 at 12:12
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    $\begingroup$ @BastamTajik: Incrementally updating questions after answers are given doesn’t fit the MathOverflow/StackExchange system very well, and is generally discouraged (questions are meant to be answerable, not a moving target) — so it’s better to accept the answer and ask the follow-up as a new question. But also, if you can see when you ask a question that it has a likely followup that fits together with it, then it’s usually best to ask them together from the start. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Oct 6, 2023 at 12:54

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