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Take some infinite hypernatural number, $M$, and consider the integers (finite and infinite) less than or equal to $M$. There are uncountably many. Then consider $\log_2 M$. Is there a straightforward way to understand the cardinality? Could it be countable?

Ludolila
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Richard S.
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If $M$ is nonstandard, so is $\log_2(M)$, so the same reasoning shows that there are uncountably many nonstandard integers $<\log_2(M)$ as well.

Noah Schweber
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    Although, I wonder: if you are in some arbitrary nonstandard model of arithmetic (not necessarily an ultrapower), is it possible that the set of numbers less than $N$ has different cardinality from the set of numbers less than $2^N$ for some nonstandard $N$? – Eric Wofsey Oct 29 '18 at 03:40
  • Starting with an infinite N, then 2^N is another nonstandard number. Is it easy to see that the sets of integral numbers less than N and 2^N have the same cardinality? Then, it seems that if you begin with N, consider the cardinality of integral numbers less than N, and consider 2 raised to that cardinality, then you'd get a larger infinity. Is that correct? – Richard S. Oct 29 '18 at 12:58
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    @RichardS. No, it's not correct. In general, actually, they won't have different cardinalities: in any sufficiently saturated model of arithmetic, we'll have $[0,M]\equiv[0,N]$ for any nonstandard $M,N$. In particular, the integers of a hyperrreal field form such a model, so - for the hyperreals at least - you'll never see different cardinalities. – Noah Schweber Oct 29 '18 at 17:09
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    And (looking at arithmetic for simplicity) there's no nonstandard model in which $[0,M]$ has smaller cardinality than $[0, 2^M]$ for every $M$, since otherwise the sequence $M,\log_2(M),\log_2(\log_2(M)),...$ would give us a strictly descending chain of cardinalities. In fact, it's not obvious - and this is what Eric is asking - that a cardinality difference can ever happen! – Noah Schweber Oct 29 '18 at 17:10
  • Thank you all for these replies--very helpful and I appreciate them a great deal. It's also becoming clear that I'm going to need to do a great deal of work if I want to think more coherently on these topics. One quick follow-up on the descending chain. Let's say a person comes along and says, "Yes, it's precisely this sort of descending chain that I am trying to get in order to prove (a fragment of) arithmetic inconsistent." One response would be to decide to ignore this person forever. But is there some other way to prove there are no such descending chains to appease this person? – Richard S. Oct 29 '18 at 17:43
  • @RichardS. First of all, this putative descending chain of cardinalities would be evidence of an inconsistency in ZFC, not any piece of arithmetic, and even then it would have to be constructible using ZFC (very weird things are consistent with ZF). Ultimately, there's no way of appeasing such a person on one's own - they need to be involved in the process too, the crucial point being that they need to recognize that the onus is on them to at least provide a heuristic argument for what they want to exist. (cont'd) – Noah Schweber Oct 29 '18 at 17:46
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    At present, the only argument I see here is "shouldn't exponentiation behave similarly on cardinalities as it does on *nonstandard numbers?," and the response is "why should it?" They are completely different types of objects. Of course this analysis takes place within a background theory, which our putative conversant is dubious of, but the point is that so far everything seems self-consistent within that system. – Noah Schweber Oct 29 '18 at 17:48