ArisKatsaris comments on Standard and Nonstandard Numbers - Less Wrong

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

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Comment author: ArisKatsaris 19 May 2013 01:15:40AM 0 points [-]

"For all x, there exists a number k such that 0 after k successions is equal to x" That should deal with another possible understanding of that infinite chain. Or is defining k in those axioms the problem?

I made roughly a similar comment in the Logical Pinpointing post, and Kindly offered a response there.

If I understood him correctly basically it meant "you can't use numbers to count stuff yet, until you've first pinpointed what a number is...". And repetition isn't defined in first order logic either.

Comment author: Indon 19 May 2013 01:51:55PM *  0 points [-]

Ah, so the statement is second-order.

And while I'm pretty sure you could replace the statement with an infinite number of first-order statements that precisely describe every member of the set (0S = 1, 0SS = 2, 0SSS = 3, etc), you couldn't say "These are the only members of the set", thus excluding other chains, without talking about the set - so it'd still be second-order.

Thanks!

Comment author: Kindly 19 May 2013 02:38:55PM *  0 points [-]

It's a bit worse than that. Even if we defined the "k-successions" operator (which is basically addition), it doesn't actually let us do what we want. "For all x, there exists a number k such that 0 after k successions is equal to x" is always satisfied by setting k=x, even if x is some weird alternate-universe number like 2*. Granted, I have no clue what "taking 2* successions of 0" means, but...