# Talk:Order topology

WikiProject Mathematics (Rated Start-class, Low-priority)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Start Class
 Low Priority
Field:  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)

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)

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

## 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 (talkcontribs)

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 ${\displaystyle \omega ^{2}\!,}$ 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 ${\displaystyle \langle (\omega \cdot n)+1\,|\,0 and every element in that sequence is mapped to one. Thus the function is discontinuous. JRSpriggs 06:44, 30 October 2006 (UTC)

## 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)

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

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)