Talk:Limit ordinal

Clarity
To Patrick: You really need to try harder to be clear. Do not assume that the reader understands what you are talking about. I am an expert on ordinals and even I have no idea what you mean by this sentence &mdash; "The set of limit ordinals of ordinal number &omega;(p + 1) is isomorphic to p; the ordinal numbers &omega;p + 1 and those in between also.". So I will remove it. Also you speak of ordinals as if they were sets, referring for example to the predecessor of an ordinal as its "maximum". While they can be viewed as sets, this is confusing. Instead of saying "&alpha;'s maximum", say "the maximum element of &alpha;" or "the largest ordinal less than &alpha;" or "&alpha;'s maximum element" for clarity. JRSpriggs 06:52, 11 June 2007 (UTC)


 * I was trying to explain "how many" limit ordinals a well-ordered set has. How about:


 * The set of limit ordinals in a well-ordered set of ordinal number ωp + n with n > 0, $$n \in \mathbb{N}$$, has ordinal number p;
 * The set of limit ordinals in a well-ordered set of ordinal number ωp has as ordinal number the predecessor of p if p is a successor ordinal, and p if it is a limit ordinal.


 * I agree that ambiguity should be avoided between the maximum of x as a set, and the maximum of a set of values of x. Adding "element" seems a good idea. Patrick 08:56, 11 June 2007 (UTC)


 * That is much better, but I would rewrite it further like this:
 * The set of limits in a well-ordered set of order type ω&middot;&rho; + n, for 0 < n < &omega;, has order type &rho;.
 * The set of limits in a well-ordered set of order type ω&middot;&rho; has order type &sigma; if &rho; = &sigma; + 1, and order type &rho; if &rho; is a limit ordinal or zero.
 * What do you think? JRSpriggs 03:37, 12 June 2007 (UTC)


 * Yes, that is fine. One thing, is it on purpose that you use "limit" and "limit ordinal"? Otherwise using the same term uniformly may be better.--Patrick 08:36, 12 June 2007 (UTC)

A general well-ordered set may not be an ordinal, and its limits may not be limit ordinals. Since you were using the expression "well-ordered set", I went with that image. Otherwise, one could say:
 * The set of limit ordinals less than ω&middot;&rho; + n, for 0 < n < &omega;, has order type &rho;.
 * The set of limit ordinals less than ω&middot;&rho; has order type &sigma; if &rho; = &sigma; + 1, and order type &rho; if &rho; is a limit ordinal or zero.

Here, instead I have used the image of ordinals as elements of an ordered class. But perhaps it would be more to the point to say:
 * An ordinal &lambda; is a limit ordinal if and only if there is an ordinal &alpha; such that &lambda; = &omega;&middot;(1 + &alpha;).

Is this better? Notice that for clarity, I try to avoid talking about ordinals as sets unless I include a word like "element" which indicates that I am talking about a set. JRSpriggs 03:56, 13 June 2007 (UTC)


 * Thanks. Maybe no addition to the article about this is needed, in the meantime I have already added earlier "It can be written in the form &omega;&alpha; for &alpha; > 0.", very similar to your last version.--Patrick 07:56, 13 June 2007 (UTC)


 * The set of limits in a well-ordered set of order type ω&middot;(1 + &rho;) + n, for 0 < n < &omega;, has order type &rho;+1.
 * The set of limits in a well-ordered set of order type ω&middot;(1 + &rho;) has order type &rho;.
 * The set of limits in a well-ordered set of order type n, for n < &omega;, has order type 0. JRSpriggs (talk) 06:13, 24 April 2012 (UTC)