TerriblySorry comments on Standard and Nonstandard Numbers - Less Wrong

31 Post author: Eliezer_Yudkowsky 20 December 2012 03:23AM

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Comment author: IlyaShpitser 20 December 2012 09:14:12AM *  21 points [-]

Now how do we get rid of the chain?

"We have to use second-order logic for that one."

No we don't.

The model with the natural numbers and a single "integer line" is not a first order model of arithmetic. The reason is this. For a non-standard number "a" large enough there is a (non-standard) natural number that's approximately some rational fraction of "a." This number then has successors and predecessors, so it has an "integer line" around it. But because we can play this game for any fraction, we need lots of integer lines (ordered according to the total ordering on the rationals).

See this for details:

http://web.mit.edu/24.242/www/NonstandardModels.pdf

Comment author: [deleted] 20 December 2012 04:42:58PM *  3 points [-]

Wikipedia concurs:

Any countable nonstandard model of arithmetic has order type ω + (ω* + ω) · η, where ω is the order type of the standard natural numbers, ω* is the dual order (an infinite decreasing sequence) and η is the order type of the rational numbers. In other words, a countable nonstandard model begins with an infinite increasing sequence (the standard elements of the model). This is followed by a collection of "blocks," each of order type ω* + ω, the order type of the integers. These blocks are in turn densely ordered with the order type of the rationals. The result follows fairly easily because it is easy to see that the non standard numbers have to be dense and linearly ordered without endpoints, and the rationals are the only countable dense linear order without endpoints.

Edit: Eliezer seems to have been aware of this, and gave a valid reply to your comment, so I won't call it a "mistake" anymore. I do think some rewording or a clarifying annotation within the OP would be helpful, though.