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Let X be a finite topological space(every pair of distinct points are topologically distinguishable). Define < to be a relation on X such that: y< x if y not equal to x, and every open set that contains y also contains x. Show that < is a strict partial order on X. How to prove x < x is false in this situation?

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It is a direct consequence of the definition. The first requirement for $y < x$ to be true is that $y \neq x$, so if $x<x$ was true, it would be a contradiction to the definition since obviously $x=x$.

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  • $\begingroup$ if y≠x and every open set that contains y also contains x than y<x, but I don't think the converse is true. Just like I cannot say that if x<x is true then x≠x and... $\endgroup$
    – kaomur
    Dec 5 '19 at 3:46
  • $\begingroup$ This depends only on the way the relation is defined. If it is defined the way you said it was in your question, then it must be true that if $x<x$, then $x\neq x$. $\endgroup$
    – Vincent
    Dec 5 '19 at 4:02
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    $\begingroup$ @kaomur: math.stackexchange.com/questions/566565/… $\endgroup$ Dec 5 '19 at 4:13
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The specialisation pre-order $\le$ is defined as $x \le y$ iff $x \in \overline{\{y\}}$, which when unpacking the definitions means "every open set that contains $x$ also contains $y$". This always obeys reflexivity ($\forall x: x \le x$) and transitivity ($\forall x,y,z: x \le y \land y \le z \to x \le z$).

A space where every pair of distinct points is distinguishable means that $$\forall x,y \in X: x \le y \land y \le x \to x=y$$

or otherwise put, $\le$ is a partial order (so also antisymmetric) instead of just a pre-order (reflexive and transitive). It turns out that this is equivalent to the $T_0$ property.

As explained on the linked Wikipedia page, we can in this case define a strict version $$x < y := (x \neq y) \land x \le y$$ and this will be a strict partial order if $\le$ is a "normal" partial order, and this is exactly what is done in your question.

By definition $x < x$ cannot hold as the left clause in the "and" ($ x \neq x$) is false, so the whole statement is false when taking $y$ equal to $x$.

There is nothing topological about the question, it's the general construction of making a strict PO out of a normal PO.

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