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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?

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    $\begingroup$ The cardinality of what? $\endgroup$ – Eric Wofsey Oct 29 '18 at 3:36
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    $\begingroup$ If $M$ is infinite, so is $\log_2(M)$ ... $\endgroup$ – Noah Schweber Oct 29 '18 at 3:37
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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.

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    $\begingroup$ 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 $N$? $\endgroup$ – Eric Wofsey Oct 29 '18 at 3:40
  • $\begingroup$ 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? $\endgroup$ – Richard S. Oct 29 '18 at 12:58
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    $\begingroup$ @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. $\endgroup$ – Noah Schweber Oct 29 '18 at 17:09
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    $\begingroup$ 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! $\endgroup$ – Noah Schweber Oct 29 '18 at 17:10
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    $\begingroup$ 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. $\endgroup$ – Noah Schweber Oct 29 '18 at 17:48

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