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)