Definition talk:Order Topology

I see your use of "mathop": how does that work? --prime mover 09:12, 27 March 2012 (EDT)

Not entirely sure, but I recalled its existence from my passive knowledge; it functions as follows (I think it's similar to the preamble function '\DeclareMathOperator'):

$\uparrow \left({s}\right)$ as opposed to $\mathop{\uparrow} \left({s}\right)$

So apparently (as the name suggests), it enforces its content to behave like an operator name (such as '\exp' etc.). --Lord_Farin 18:05, 27 March 2012 (EDT)

... in practical terms, it appears to adjust spacing accordingly. Works for me, then. --prime mover 18:07, 27 March 2012 (EDT)

Equivalent Definition

I'm mostly putting this here to remind myself to do it, but I have a slightly different (it seems like it's obviously equivalent) definition of an order topology from Munkres:

Let $X$ be a set with a simple order relation and more than one element. Let $\mathscr{B}$ be the collection of all sets of the following types: All open intervals $(a .. b)$ in $X$, all intervals of the form $[a_0 .. b)$ where $a_0$ is the smallest element if any of $X$, and all intervals of the form $(a .. b_0]$ where $b_0$ is the largest element if any of $X$. Then $\mathscr{B}$ is the basis for the order topology on $X$. --Alec (talk) 12:30, 14 April 2012 (EDT)

Yes, that's entirely equivalent. The upper and lower closures are resp. the third and second types, and the first type arises by intersection of these. I'm quite happy that my self-conceived definition actually is present somewhere in the literature. --Lord_Farin 13:22, 14 April 2012 (EDT)