Talk:Successor of Omega

The name of this page needs to be shorter. --prime mover 14:32, 29 January 2012 (EST)

This seems to apply more to ordinals rather than to the natural numbers. That is, to $\omega$ rather than $\N$. Furthermore, $\N+1$ can easily be read as set-wise addition, yielding it equal to $\N \setminus \{0\}$... --Lord_Farin 03:19, 30 January 2012 (EST)

You are correct, of course. I elected to use $\mathbb N$ rather than $\omega$ because most people fail to realize that $\omega = \mathbb N$. Using the symbol $\omega$ is actually better provided one is certain that the reader understands this important fact. --KBlott

The area of ordinals on PW appears to lie (mostly) in darkness. I have a reasonable resource (lecture notes; I know the author) to back up most stuff. The equality $\omega = \N$ holds set-wise. However, as you point out, $\omega$ usually denotes the ordinal, while $\N$ also subsumes the additive structure and all the like. I think this correspondence is already up here on Definition:Minimal Infinite Successor Set, Natural Numbers as Successor Sets and whatever is linked from there. --Lord_Farin 14:29, 30 January 2012 (EST)

Considering the $+1$ notation: $A+1$ is indeed common, but on PW it is written $A^+$, to avoid the confusion encountered earlier. It can be found on referenced pages. --Lord_Farin 15:07, 30 January 2012 (EST)

What does + 1 mean?

The essential point that people need to know to understand this concept is

$ \varnothing =: 0 $

$\in \lbrace 0 \rbrace =:1$

$ \in \lbrace 0, 1 \rbrace =: 2$

$ \in \lbrace 0, 1, 2 \rbrace =: 3$

…

$ \in \lbrace 0, 1, 2, …, n \rbrace = :n+1$

…

$ \in \lbrace 0, 1, 2, … \rbrace =: \mathbb N$

$ \in \lbrace 0, 1, 2, … ; \mathbb N \rbrace$ (Copi used the semicolon to indicate that $\mathbb N$ is not part of the sequence $0, 1, 2, … $)

$=: \mathbb N + 1$

$ \in \lbrace 0, 1, 2, ... ; \mathbb N, \mathbb N + 1 \rbrace =: \mathbb N + 2$

...

In other words, for every ordinal number $x$, $x + 1 = x \cup \lbrace x \rbrace$. --KBlott 12:13, 30 January 2012 (EST)

...But what does $+$ mean? --GFauxPas 12:50, 30 January 2012 (EST)

For every ordinal number $x$, $x + 1$ means $x \cup \lbrace x \rbrace$ by definition. As Lord_Farin correctly observed there is a need to disambiguate$\ + 1$ operating on an ordinal number from$\ + 1$ operating on a set. Otherwise one would be lead to the conclusion that

$\lbrace \mathbb N, 0, 1, 2, ... \rbrace =\lbrace 0, 1, 2, ... ; \mathbb N \rbrace = n \cup \lbrace n \rbrace = \mathbb N + 1 = \lbrace 1, 2, 3, ...\rbrace$.

That, in turn, would lead us to conclude that

$\lbrace 0 \rbrace \cup \lbrace \mathbb N \rbrace = \lbrace 0, \mathbb N \rbrace = \varnothing$.

The problem here is that ordinal numbers $are$ sets! -KBlott

Okay this stuff is above my head, but other readers will be above their heads as well. This stuff you're introducing needs to be defined/linked to on the page, like what an ordinal number is, and what addition of ordinal numbers is, and what a semicolon is. And the fact that ordinal numbers are sets, well, most/all numbers are sets anyway, once you get to the nitty gritty, no? --GFauxPas 13:40, 30 January 2012 (EST)

Makes sense to me. (Just think of the semicolon as a comma if the notation gets in the way of comprehension.) --KBlott 13:53, 30 January 2012 (EST)

What's the big deal anyway?

I've just covered all this when I worked my way through the first 14 chapters of Halmos over the last few weeks. (This page may not have been posted up quite yet, but all the supporting work has been posted up, and the language and notation ought to match what has gone before.)

It's all very well citing a notation or a definition, but unless there is a link to a page somewhere else on ProofWiki that actually tells you what it means, it is of little use.

If the use of the semicolon is the same as a comma, then there's no point to it and a comma should be used instead, otherwise it will confuse everyone who will say: "What does the semicolon mean?" If it is not the same as a comma, then explain what it means. If, as you suggest, "Copi wouldn't like this", then Copi is going to have to complain in person. Except he can't because he's dead. --prime mover 16:49, 30 January 2012 (EST)

House style

With my limited understanding of transfinite arithmetic I tried to change this to house style. Please proofread. --GFauxPas 09:10, 31 January 2012 (EST)

The information appears to be correct, except for the links to pages not yet created and the reference to Copi. The style in the eqn template is not what I would use, but I wouldn't want to enforce my standard as 'the undeniably most proficient and clear' on PW. Now that the contents of the page have crystallised, it strongly feels like an example rather than a full-fledged theorem; but that's just my opinion. It might be time to finally start developing the Example namespace, but I feel that might be rushing it; a lot of other work is waiting as well. I will put it on my long-term projects list. --Lord_Farin 10:38, 31 January 2012 (EST)

Semicolon

Funny I thought I'd posted this ...

It appears it probably does make sense to use the semicolon notation because $\{0, 1, 2, \ldots; \N\}$ is subtly different to $\{\N, 0, 1, 2, \ldots\}$. We need to explain it, but that is a detail. Explanation will probably go in Definition:Set or one of its offshoots. --prime mover 12:51, 31 January 2012 (EST)

All finished

Right, this page is now as good as I can get it. The semicolon notation has been (indirectly) explained, and the $\omega$ notation is used as it ought to be. Happy trails. --prime mover 14:41, 31 January 2012 (EST)