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Comment by godelski | Hacker News Reader

It always made more sense to me to define things as "one to one AND that one set's of elements is not absolutely contained in the other set." This would make the integers bigger than the even integers because 2Z in Z but would mean that the evens and the odds are the same size, since they are disjoint sets. Though both are obviously categorically countable.

But I'm not a (set) theorist and I'm sure this has been explored before. Does anyone have a link and/or concise explanation? (papers are fine) Would this definition result in similarity weird conclusions? (Infinities are weird)

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Consider the following sets: the primes (let's call it P), the odd numbers (let's call that O), the odd numbers except for 3 (and let's call this O3).

Using your definition O would be equal in size to P, P would be equal in size to O3, but O3 would be smaller in size than O.

Ah thanks! That's an obvious counter example!

But as a followup, are there definitions that rely the rate at which sets approach infinity? P clearly "fills" its set more slowly than the even integers whereas the evens and odds "fill" the set in similar times. This would obviously mean 2Z, 3Z, and 3Z - {3} would be the same size (unless we invoke the disjoint requirement), but these could be used categorically like Big O notation (which can be refined).

Would this an even useful metric? Are set theorists even interested in differentiating these infinities?

Edit: I also gave a bit more motivation in the reply to your sibling comment. Short is that if we do "Z - 2Z" we can get the two sets of positive odds and negative odds without {0}. It seems reasonable that since we can do this decomposition and match in the normal manner that since there is a remainder that one would be larger than another but this also does not clear up the example you provided with primes.

I always thought that a "growth rate" definition of sizes of infinity makes more sense than Hume's principle. Then the odd numbers approach infinity half as fast as the natural numbers. But that only works for ordered quantities.

I mean this doesn't actually solve the issue: 2Z and 3Z are the same size under this definition.

> 2Z and 3Z are the same size under this definition

That's quite the intention actually. The motivation here is that there is clearly a 1-to-1 and onto (bijective) mapping between these sets. While we can set a clear map even from Z to 2Z (T: x -> 2x) there are other maps that work perfectly as well. For example if we take the ideal/subset of Z such that we use every other number (i.e. the evens) then this also corresponds to 2Z but we still have an infinite number of numbers left over.

Let's be more clear

Z = {Z+} + {Z-} + {0} = {2Z+} + {(2n+1)Z+} + {2Z-} + {(2n+1)Z-} + {0}

2Z = {2Z+} + {2Z-} + {0}

Then if we do Z - 2Z we get {(2n+1)Z+} + {(2n+1)Z-} (or the odds without 0). In other words, we have infinity plus infinity but one of the infinities is fully contained in the other infinity. It does seem reasonable to think that these may not be the same size, just as we would say that the uncountable infinities is larger than the countable infinities. In fact, the logic is quite similar here.

Note though, that the sibling comment (to you) gave a counter example as to where there might still be issues. Because by this setup we'd still have the number of primes equal to the size of even integers. So I asked about rates of going to infinity.

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