# strict partial ordered set and topologically distinguishable

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?

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

• 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... Dec 5 '19 at 3:46
• 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 $x\neq x$. Dec 5 '19 at 4:02
• Dec 5 '19 at 4:13

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.