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Let $(X,\leq)$ be a preorder, i.e. $\leq$ is a reflexive and transitive binary relation on $X$. I want to show that $\leq$ induces a topological structure on $X$. Hence I need to specify when a subset $A$ of $X$ must be considered open. My personal guess is that $A$ is open iff it is contained in its interior, where the interior $A^°$ of $A$ is defined as follows:

$$A^°:=\{s\in X: \forall f\in X (s\leq f\rightarrow f\in A\}$$

Do you think this makes sense? If not, which could be the right definition of an open?

A "dual" definition for the interior of $A$ could be

$$A^°:=\{s\in X: \forall p\in X (p\leq s\rightarrow p\in A\}$$

Do the two definitions work in order to get a topological space? Do they define the same topology on $X$ (if they define any?)

In the same spirit, we could define a topology by mean of closed subsets. I say that a subset $C$ of $X$ is closed when the closure of $C$ is contained in $C$, and i define the closure of $\overline{C}$ of $C$ as follows: $$\overline{C}:=\{s\in X:\exists f\in X(s\leq f\wedge f\in C\}$$ or, dually, as $$\overline{C}:=\{s\in X: \exists p\in X(p\leq s\wedge p\in C\}$$ The question is the same: do this define a topology on $X$? If so, what kind of topology is this? Has it some evident description, is it the obviuous one, or natural in some sense?

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I don't understand the notation in your question. To show that every preorder defines a topology why not specify a basis and then consider the topology generated by the basis? –  user38268 May 27 '13 at 7:53
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Every space is a topological space. You want to specify that there is a topological structure on $X$ compatible with the preorder (i.e. every order preserving endomorphism is continuous, or something like that) –  Asaf Karagila May 27 '13 at 7:55
    
@BenjaLim yes, it could be a way, but i can't do that. How to define a basis? –  Federica Maggioni May 27 '13 at 7:56
    
Your definition of interior will often return the empty set. For instance, if we give $\mathbb{R}$ the usual order, then the interior of $(a,b)$ under your definition is empty, since given any $s$, we can find $f$ such that $s \leq f$ and $f \notin (a,b)$. –  Zach L. May 27 '13 at 7:56
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There are a trilion different ways to turn a set into a topological space. Without further specifying some conditions relating the topology and the given pre-order there is no hope at finding any particular topology among the many that can be entertained. You can't prove that a pre-ordered set is a topological space since it is not. It may induce a topology. That is a different matter. –  Ittay Weiss May 27 '13 at 7:58

2 Answers 2

If you have a simple order, then the following approach by Munkres is a nice one:

The following will be the basis elements of the topology, and the generate the topology using this basis.

  1. All intervals of the form (a,b) where $a,b \in X$.
  2. All intervals $(a_{0},b)$ where $a_{0}$ is the smallest element (if any) of the space $X$.
  3. All intervals $(a,b_{0})$ where $b_{0}$ is the largest element (if any) of the space $X$.

It is easily seen that this forms a basis of a topology.

This is one way to define a topology on an ordered set. For example, the usual topology on $\mathbb{R}$ is the one obtained by this method.

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Trying to enter into the mind of the OP I think she wants to mimic the usual euclidean topology on $\mathbb{R}$: so you can define a basis for your topology given by $ \{ (a,b):a,b\in X \}$. You have to check that it satisfies the condition for being a basis and if it's true or not that order preserving endomorphism are continuous with this topology on both domain and codomain.

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