I'm interested in exploring whether there is a monad at infinity. I guess we would define the infinitesimal space surrounding infinity as "A number that is greater than any Real number, but smaller than infinity". I can see some problems with it though - there seems to be no way to differentiate between infinity and near-infinity, since if omega = infinity, then omega - a = omega for all a an element of the reals. At the same time any object that is greater than the reals but smaller than infinity must be infinitesimally close to infinity. But Infinity is an attractor in this sense, in that all numbers that are within any commensurable distance of infinity must be infinite in magnitude. So there cannot be a monad of infinity? Am I even close?
2 Answers
This question seems based on a misconception.
There is not a single thing called "infinity" in the$^1$ hyperreals. Rather, there are many different hyperreals which are infinite and quite distinct from each other. Put another way, there is no "distinguished" infinite hyperreal, any more than there is a "distinguished" infinitesimal hyperreal.
Meanwhile, by definition a hyperreal is infinite iff it is greater than every real number, so trivially there's nothing between finite and infinite in the positive hyperreals.
$^1$Actually, it's an abuse of terminology to even talk about "the hyperreals" - there is no specific thing called the hyperreals. Rather, there is a notion of "hyperreal field," and we can prove that lots of genuinely different (= non-isomorphic) hyperreal fields exist.
Generally in nonstandard analysis it doesn't matter which we use, so we simply pick one "in the background" and call it the hyperreals, but this is an important subtlety worth mentioning. (And I believe there are more technical situations where we actually do care, to a certain extent at least, which hyperreal field we use - although I can't find an example of this at the moment.)
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Thanks for your reply, Noah. I was merely trying to put some symmetry around the definition of infinitesimals for my own understanding. Thanks also for clarifying that there is no "distinguished" infinitesimal hyperreal.I understand that "R* has a positive infinitesimal, that is, an element e such that 0 < e and e < r for every positive r an element of R." --FOUNDATIONS OF INFINITESIMAL CALCULUS (H. Jerome Keisler)-- my reversal of the inequality for "numbers near infinity" above seemed justifiable to me. – Spanki Feb 08 '20 at 00:35
The problem is that you see infinity on the same axis as normal numbers, leading you to believe the “last” number of normal numbers and the “first” number of that new axis “touch”.
Try seeing those numbers like a new axis. I mean i gets its own axis for much less hefty reasons.
Then in becomes trivial to resolve paradoxes regarding infinity.
And don’t be discouraged by the cultists from the non-scientific branch of mathematics getting triggered by you daring to “offend” their dogma err I mean “axiom”. ^^
A further hint is to think about ”partial” numbers. Think partial functions, but for numbers. 5/0 never needs to be fully resolved to get across the singularity called infinity, if you allow for a new axis to “remember” 5 across it.
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Thanks for that! I really liked your suggestion of using a new axis! Helps a lot!
Now I'd like to try to understand how one can determine that one infinitesimal is demonstrably different from another - if that even makes sense. To me it seems that an infinitesimal, e_1, is different in value from another infinitesimal say e_2, when e_1 - e_2 != 0. But these values are smaller than any real. So how do we show that they are different? Or am I way off base here? is there in fact only one infinitesimal?
– Spanki Sep 29 '22 at 20:16 -
@Spanki: Well, just like you would show that one complex number is different from another complex number, when only the imaginary part differs. That is what I meant with “remember”. – Oct 04 '22 at 07:25
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So we could create a number such as (x + y*e) where x, y are elements of R! Thanks. But what about multiplying infinitesimals? Say we have e^n (n is in N). Is there a cut off point that is too close to 0 to be meaningful? E.G e^((10^10)^100)? – Spanki Oct 05 '22 at 16:54