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Let $X$ denote a set and consider a collection $C \subseteq \mathcal{P}X$. Let $\mathcal{T}$ denote the set of all topologies $T$ on $X$ such that $C \subseteq T$. Then $\bigcap \mathcal{T}$ is a topology on $X$. In particular, its the least topology on $X$ that includes $C$. So every collection $C \subseteq \mathcal{P}X$ generates a topology in this way.

Now let $(P,<)$ denote an (irreflexive) partially ordered set. Lets define that $(a,b) = \{x \in P | a < x < b\}$. Let $C=\{(a,b)|a,b \in P\}$. Then $C$ generates a topology in the sense outlined above. Is $C$ necessarily $T_0$?

More ambitiously, let $(P,\precsim)$ denote a quasiorder, and define a relation $<$ by asserting that $a<b$ precisely when $a \precsim b$ but not $b \precsim a$. Then $(P,<)$ is an (irreflexive) partially ordered set. Now repeat the above paragraph: define $(a,b) = \{x \in P | a < x < b\},$ etc. Is it true that the topology generated in this way is $T_0$ if and only if $\precsim$ is in fact a partial order?

Intuitively, the answer is "yes," because a partially ordered set can be obtained from a quasiordered set by taking a quotient. And the Kolmogorov quotient of a topological space is always $T_0$. So intuitively, we've already taken a quotient, and shouldn't have to take another quotient.

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Something is wrong with my answer? – Asaf Karagila Feb 25 '13 at 1:39
No I changed accidentally. I'm asking another question on this topic, and I considered rephrasing this one, but instead will post a new one. – goblin Feb 25 '13 at 1:49
up vote 2 down vote accepted

What happens when you apply this to the following poset:

 /   \
b     c
 \   /

Suppose that $(P,<)$ is a partially ordered set (irreflexive and transitive) such that the order topology is $T_0$, what can we conclude on the structure of $P$?

It means that given two points there is an interval containing one and not the other, namely given $x,y\in P$ if there is some $z\in P$ such that $z<x,y$ then there is $z'$ such that exactly one of the following occurs, $x<z'$ or $y<z'$.

But this is still not enough, because if there are two isolated points (order-wise) then they are not contained in any such interval. So we need to require that the order itself contains no more than one point without neighbors.

But this is an ugly condition, we may lose exact bound on $T_0$-ness, but we can give the following statement:

Let $(P,<)$ is a partially order such that:

  1. For every $x$ there is some $u,v$ such that $u<x<v$; and
  2. For every $x,y$ if there is some $z$ such that $z<x$ and $z<y$ then there is some $z'$ such that $x<z'$ or $y<z'$ but not both.

Then the order topology on $P$ is $T_0$.

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Good point Asaf. See above. – goblin Feb 3 '13 at 19:44
Do you mean to ask what sort of requirements on the order are there for the order topology to be $T_0$? – Asaf Karagila Feb 3 '13 at 19:46
Yes, precisely. – goblin Feb 3 '13 at 19:47
Good answer. When you speak of the order topology, does this include the open rays? – goblin Feb 3 '13 at 20:54
Well, if the order is linear then the open rays are included because they union of increasing intervals; if the order is not linear what does it mean "open rays"? Do you mean "cone topology" (i.e. the open sets generated by $L_x=\{y\mid x<y\}$ or something like that)? – Asaf Karagila Feb 3 '13 at 21:04

No. For example, let $X$ be a set with more than one element, and let $<$ be the trivial partial order: that is, $x<y$ is always false. The the induced topology is the trivial topology (that is, the only open sets are the empty set and $X$ itself) which is not $T_0$.

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Oh good point. Any idea whether the concept can be salvaged? – goblin Feb 3 '13 at 19:43
One option (which is fairly standard) is to (a) restrict to total orders and (b) allow intervals of the form $(a,\infty)$ and $(-\infty,b)$. Then your order topologies are always Hausdorff. – Chris Eagle Feb 3 '13 at 19:46
Yes. According to wikipedia, restriction to total orders results in a completely normal Hausdorff space. I'd be happy with $T_0$, but without the restriction to total orders. Adding in the open rays makes sense. – goblin Feb 3 '13 at 19:49

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