Talk:Order topology

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 Field:  Topology

RazorFlame undid correction[edit]

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)

Previous comments[edit]

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"?[edit]

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[edit]

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 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 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?[edit]

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 .

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

Definition is too restrictive[edit]

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)