Talk:Order topology

RazorFlame undid correction

Right order and left order are REVERSED. I fixed it and some tool/bot Razorflame reverted them. —Preceding unsigned comment added by 140.247.149.65 (talk) 21:08, 24 October 2009 (UTC)[reply]

Previous comments

show that IxI in the dictionary order topology is locally connected but not locally path connected.what are the components of this space?

I don't understand IxI. Do you mean the cartesian product of 2 intervals?

MFH 17:30, 8 Apr 2005 (UTC)

Move section here from "Ordinal number"?

If no one raises an objection in the next few days, I will move the section Ordinal number#Topology and ordinals from that article to this one. My objective is to make that article shorter (and this one longer). JRSpriggs 05:22, 25 September 2006 (UTC)[reply]

Section moved in. JRSpriggs 08:33, 30 September 2006 (UTC)[reply]

Continuous functions from omega-1 to the reals

I think there should be a modification:-

It is also worthy of note that any continuous increasing function from ω₁ to R (the real line) is eventually constant

Because if it just has to be continuous I can define it to be 1 at successors of limit ordinals and 0 elsewhere. —Preceding unsigned comment added by 81.179.130.150 (talk • contribs)

No. The original formulation, "... any continuous function from ω₁ to R (the real line) is eventually constant ...", is true. The function you specified is not continuous. Consider

\omega ^{2}\!,

it is a limit (rather than the successor of a limit) so you give it the value zero. However, it is the limit of the sequence

\langle (\omega \cdot n)+1\,|\,0<n<\omega \rangle

and every element in that sequence is mapped to one. Thus the function is discontinuous. JRSpriggs 06:44, 30 October 2006 (UTC)[reply]

subspace topology finer than induced order topology?

It says: "subspace topology is always finer than the induced order topology;" I think that this is in the wrong direction though; i.e. it's never finer, but could be coarser. I'm not completely sure, but Section 4.1 problem 9 on page 118 of Folland's Real Analysis asks you to prove this; and I'm pretty sure my proof is correct, although it's somewhat messy. I don't trust the example given after it either; but I think it would be better if somebody more acquainted with the subject looked it over rather than me messing with it.

Greeneggsnspam (talk) 08:28, 14 November 2008 (UTC)[reply]

The example seems to be messed up. I think the author meant the set $\{0\}\cup \{1/n:n\in \mathbb {N} \}$ .

Anonymous 12:42, 12 March 2011 (UTC) —Preceding unsigned comment added by 89.27.245.109 (talk)

Definition is too restrictive

Why the definition is restrictive to totally ordered sets? The order topology for partially ordered sets is a nice thing to build interesting counterexamples. Albmont (talk) 18:32, 30 April 2009 (UTC)[reply]

Sections on ordinals

There are multiple sections regarding ordinals, which have some overlap. It would be worthwhile to merge these sections and remove redundancies. --Jordan Mitchell Barrett (talk) 05:26, 2 May 2020 (UTC)[reply]

When X is singleton

The problem with the case when $X=\{x\}$ is a singleton, is that we do not get a base/subbase. We only can choose $a=b=x$ and we get $(-\infty ,b)=(a,b)=(a,\infty )=\emptyset$ . So this system is not cover of X, which is one of necessary conditions of a base. (And in many sources, this is required also form a subbase.) Although this special case is probably not important enough to be explicitly mentioned. --Kompik (talk) 20:09, 4 November 2020 (UTC)[reply]

Maybe one way to solve it would be to use the sets (a,M] and [m,b) where

m=\min X

and

M=\max X

. However, this only works if the set has smallest/largest element. So with this approach, we would not need have singleton as a separate case. But the disadvantage is that the definition would be split into several cases, depending on whether the set $X$ has a smallest/largest element. --Kompik (talk) 05:25, 5 November 2020 (UTC)[reply]

I agree that there is a problem: the topology defined by the rays and the topology defined by the open intervals are different, since, for example, in $\{0,1\}$ , the first is the discrete topology and the second is the undiscrete topology. — Preceding unsigned comment added by 2001:861:3743:6E00:47F:FC94:30B1:29FF (talk) 12:58, 17 December 2021 (UTC)[reply]

Example section is opaque

The section titled An example of a subspace of a linearly ordered space whose topology is not an order topology is utterly opaque. I re-read it several times and keep getting stuck at the sentence Since {-1} is open in Z, there is some point p in M such that the interval (-1, p) is empty. I think this is trying to appeal to some separation axiom, but, well, I can't really tell. This section was added by a single user in 2006, and it is that user's only contribution to wikipedia. I dunno. There must be some easier way to state this example... 67.198.37.16 (talk) 06:28, 27 November 2023 (UTC)[reply]