Talk:Ordinal arithmetic

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[edit] Creation

I created this article by extracting the section "Arithmetic of ordinals" from the article "Ordinal number". JRSpriggs 09:34, 15 March 2006 (UTC)

[edit] Warning about notation

I inserted the following sentence that was removed

"In practice, unless it is clear from context that an expression such as 2ω is intended to denote ordinal exponentiation, it is read to mean the corresponding cardinal exponentiation (in this case, 2^{\aleph_0})."

This was motivated by a discussion here. What is the objection? Is the concern about the other possible meanings of 2^ω? CMummert · talk 13:50, 31 March 2007 (UTC)

I just disagree with your statement (especially in an article on ORDINAL arithmetic) that one should presume that 2ω really means 2^{\aleph_0}. I always presume that it means what it says, i.e. the ORDINAL exponential. JRSpriggs 06:40, 1 April 2007 (UTC)
In my field, 2^ω is used either to denote 2^{\aleph_0} or to denote the set of functions from 2 to ω; the number of times I have seen it used to indicate ordinal exponentiation could be counted on my fingers. That's why the "unless it is clear from context" phrase. CMummert · talk 12:23, 1 April 2007 (UTC)
I think you meant functions from ω to 2. Of course, context will make it clear in almost all cases. But if other aspects of the context do not make it clear, then the remaining part of the context, i.e. whether the symbols being used are symbols for ordinals or for cardinals, should be allowed to decide.
I noticed that you started to remove the italics from the Greek letters. I am now aware that not using italic Greek letters is the rule here. But when I worked on this article over a year ago, I thought that it was the other way around. So most of the Greek letters are italicized. Do you think that we should bother to remove all the italics from Greek letters in this article and also in ordinal number and large countable ordinal? JRSpriggs 06:27, 2 April 2007 (UTC)
No, I don't think its worth it. I had second thoughts after editing it that I shouldn't have removed the italics, since the entire article uses them. I'll put them back, so the article is self-consistent. CMummert · talk 11:50, 2 April 2007 (UTC)
Is this related to the notation ω2, which I've seen in regards to functions (mappings) of sets? — Loadmaster 21:30, 11 June 2007 (UTC)
Yes. Some people use ω2 and some use 2ω for the set of functions. The cardinality of that set is 2^{\aleph_0}. If you restrict the functions to those with finite support, then its order type is the ordinal 2ω. JRSpriggs 05:18, 12 June 2007 (UTC)

[edit] Cantor normal form

The article says:

Every ordinal number α can be uniquely written as \omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k where k is a natural number, c_1, c_2, \ldots, c_k are positive integers, and \beta_1 > \beta_2 > \ldots > \beta_k are ordinal numbers

and also

ωβc + ωβ'c' is ωβ'c' if β' > β

Does that not imply that

\omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k = \omega^{\beta_1} c_1

 ? --Michael C. Price talk 12:05, 5 October 2007 (UTC)

The first equation is for ordinals, but the second is for cardinals. For example ω + ω is not ω in the sense of ordinal addition. — Carl (CBM · talk) 13:23, 5 October 2007 (UTC)
If that is the case the article needs a major check over, since the 2nd equation reads in full:
The Cantor normal form also allows us to compute sums and products of ordinals: to compute the sum, for example, one needs merely know that ωβc + ωβ'c' is ωβ'c' if β' > β
which specifies "ordinal" not "cardinal". And ω is the symbol for an ordinal. --Michael C. Price talk 13:29, 5 October 2007 (UTC)
There is also the commutativity issue; I misunderstood where your confusion was and didn't read far enough in the article. Ordinal addition is not commutative. For example \omega + \omega \cdot \omega is ω followed by ω copies of ω, which is the same as just \omega \cdot \omega, so the sum collapses. But if you put the sum in the other order, \omega \cdot \omega + \omega then it does not collapse. In the article text, the Cantor normal form is set up so that the exponents decrease from left to right, so that the sum does not collapse. When you add two ordinals in Cantor normal form, you can simplify that sum by collapsing binary sums with the first term of the second ordinal. — Carl (CBM · talk) 13:55, 5 October 2007 (UTC)
Ah! I overlooked the commutativity issue. I shall reread it. Thanks. --Michael C. Price talk 14:16, 5 October 2007 (UTC)

Another question. I'm okay with:

\alpha n = \omega^{\beta_1} c_1 n + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k}c_k, if n is a non-zero natural number.

but doesn't that imply

\alpha\omega = \omega^{\beta_1+1} + \omega^{\beta_2} c_2 + \cdots + \omega^{\beta_k}c_k.

rather than just

\alpha\omega = \omega^{\beta_1+1}.

? i.e. what is the mechanism that cancels the extra RHS terms in the limit as n -> ω ? (Does n have to be in the final term in the sum for this limit substitution to work?) --Michael C. Price talk 11:55, 10 October 2007 (UTC)

So \alpha \omega = \sup \{ \alpha n : n < \omega\}. Since \alpha n < \omega^{\beta_1 + 1} for each n, this means \alpha \omega \leq \omega^{\beta_1+1}. But also \omega^{\beta_1 + 1} = \omega^{\beta_1}\omega = \sup \{ \omega^{\beta_1}n : n < \omega\}. Since \alpha n \geq \omega^{\beta_1}n for all n < ω, this means that \alpha \omega \geq \omega^{\beta_1+1} (take the sup of both sides). Since both inequalities hold, \alpha \omega = \omega^{\beta_1+1}. I'm not sure if there is a good way to visualize the infinite product, except to think of the fact that any time there is a copy of ωβ + 1 in the product it will absorb all copies of ωγ for γ < β that appear to its left. In the product αω everything in the infinite product is followed by at least one copy of \omega^{\beta_1}, so everything gets absorbed. — Carl (CBM · talk) 14:05, 10 October 2007 (UTC)
Thanks, that was most helpful. --Michael C. Price talk 09:42, 11 October 2007 (UTC)