Clearly every single rational number is "useful", plus others that are not ratio...

perl4ever · on March 25, 2020

So I guess what I lack is an understanding of why that doesn't affect cardinality.

shawnz · on March 26, 2020

There is a 1-to-1 mapping between the sets, so by definition that means they have the same cardinality.

Let me give a more common example. Consider these two sets: N, the set of non-negative integers, and Z, the set of all integers. Clearly, everything in N is also in Z, and then some. But N and Z still have the same cardinality, because there are 1-to-1 mappings between the two sets. Here is one example of such a mapping:

    N  |  Z
   ---------
    0 ->  0
    1 ->  1
    2 -> -1
    3 ->  2
    4 -> -2
    5 ->  3
    6 -> -3

..etc. The formula for this mapping would be floor(n/2)*(-1)^(n%2). Clearly everything in the left set has exactly one corresponding item in the right set and vice versa, so they must be the same "size", even though the right set contains every item in the left set and then some.