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Let $(X,\leq)$ be an ordered space and define on $X$ the order topology form the subbasis $\{(x,y), (x,\to), (\leftarrow,x) : x,y \in X \}$.

Can you read the fact that $\leq$ is a total order on the topology of $X$?

I hoped it was equivalent to a separation property of $X$ (for example, if $\leq$ is total then $X$ is Hausdorff, but there exist Hausdorff partial ordered spaces; a more difficult result is that a linearly ordered space is completely normal, and here I didn't find a counter-example).

Do you know something about that?

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up vote 3 down vote accepted

No, you cannot: the same topology can be induced both by a linear order and by a non-linear partial order. Specifically, the discrete topology on a set of any cardinality greater than $3$ can be induced both by a linear order and by a non-linear partial order (and it’s hereditarily perfectly normal and hereditarily collectionwise normal, too!). As usual, I’m assuming the axiom of choice.

Let $S$ be any non-empty set, let $X=S\times\{0,1\}$, and for $\langle s,i\rangle,\langle t,j\rangle\in X$ let $\langle s,i\rangle\le\langle t,j\rangle$ iff $s=t$ and $i\le j$. Then the induced order topology on $X$ is the discrete topology, which is hereditarily perfectly normal and hereditarily collectionwise normal. Every infinite discrete space can be obtained in this way, as can every finite discrete space of even cardinality. For finite spaces of odd cardinality just add a third point $\langle s,2\rangle$ to one of the pairs $\langle s,0\rangle,\langle s,1\rangle$ with the obvious extension of the order. For $|X|>3$ the order is non-linear.

The discrete topology on any set can also be induced be a linear order. This is obvious for finite sets. For $\kappa\ge\omega$ let $X=\kappa\times\Bbb Z$; the lexicographic order on $X$ is then a linear order that induces the discrete topology on $X$, whose cardinality is $\kappa$.

(Nice question, by the way.)

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