5 ms·

This is an interesting idea but I don't get how it works in practice. For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper s

This is an interesting idea but I don't get how it works in practice. For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper subset of the other. Are you suggesting these be treated as "the same size as" one another? Because if so, then {1,2,3,4} is "the same size as" both {A,B,C} and {A,B,C,D}, even though {A,B,C} is not "the same size as" {A,B,C,D}. On the other hand, we can say that although {A,B,C} is not equal to any proper subset of {1,2,3,4}, it is in bijection with a proper subset, such as {1,2,3}. Although this is true, if we use this definition of "the same size as," we again get that the naturals are "the same size as" the set of all squares.

bheadmaster 3y ago
> For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper subset of the other. The existence of an injective function is what makes subset {A,B,C} a "proper subset" [sic] of {1,2,3,4}, in context of comparing their size. The terminology could be chosen better, but I think the intended meaning is clear.
- ComplexSystems 3y ago
  This is exactly what I wrote about in my second paragraph, isn't it? The intended meaning is not clear, because this definition again makes the set of naturals the same size as the set of squares. > On the other hand, we can say that although {A,B,C} is not equal to any proper subset of {1,2,3,4}, it is in bijection with a proper subset, such as {1,2,3}. Although this is true, if we use this definition of "the same size as," we again get that the naturals are "the same size as" the set of all squares.
  - bheadmaster 3y ago
    Ah, I see. I have't read your comment properly, I guess. That's a fair point, it really seems that comparing sets in general only makes sense in terms of bijections, which makes infinite sets as comparable as finite sets.
cubefox 3y ago
> For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper subset of the other. Are you suggesting these be treated as "the same size as" one another? No. The paradox is that intuitively 1. If there is bijection between F and G, they are the same size. 2. If F is a proper subset of G, F and G are not the same size. For finite numbers, these principles are compatible, but for infinities, it can happen that they contradict each other.
- ComplexSystems 3y ago
  Ok. So are you suggesting a notion of "size" such that A is a smaller "size" than B if and only if A is extensionally equal to a proper subset of B? This would lead to a partial order on all sets. As a result, the perfect squares would smaller than the naturals, but also it would make many sets incomparable in size, including the natural numbers and the set {A}. Or if this is not what you are suggesting, what are you suggesting?
  - cubefox 3y ago
    I find Galileo's conclusion plausible that neither principle holds for infinite sets.

bheadmaster 3y ago

> For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper subset of the other. The existence of an injective function is what makes subset {A,B,C} a "proper subset" [sic] of {1,2,3,4}, in context of comparing their size. The terminology could be chosen better, but I think the intended meaning is clear.

ComplexSystems 3y ago
This is exactly what I wrote about in my second paragraph, isn't it? The intended meaning is not clear, because this definition again makes the set of naturals the same size as the set of squares. > On the other hand, we can say that although {A,B,C} is not equal to any proper subset of {1,2,3,4}, it is in bijection with a proper subset, such as {1,2,3}. Although this is true, if we use this definition of "the same size as," we again get that the naturals are "the same size as" the set of all squares.
- bheadmaster 3y ago
  Ah, I see. I have't read your comment properly, I guess. That's a fair point, it really seems that comparing sets in general only makes sense in terms of bijections, which makes infinite sets as comparable as finite sets.

cubefox 3y ago

> For instance, the two sets {1,2,3,4} and {A,B,C} are such that none is equal to a proper subset of the other. Are you suggesting these be treated as "the same size as" one another? No. The paradox is that intuitively 1. If there is bijection between F and G, they are the same size. 2. If F is a proper subset of G, F and G are not the same size. For finite numbers, these principles are compatible, but for infinities, it can happen that they contradict each other.

ComplexSystems 3y ago
Ok. So are you suggesting a notion of "size" such that A is a smaller "size" than B if and only if A is extensionally equal to a proper subset of B? This would lead to a partial order on all sets. As a result, the perfect squares would smaller than the naturals, but also it would make many sets incomparable in size, including the natural numbers and the set {A}. Or if this is not what you are suggesting, what are you suggesting?
- cubefox 3y ago
  I find Galileo's conclusion plausible that neither principle holds for infinite sets.