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I know that common sense is to define set $A$ to be equivalent to a set $B$ if and only if there exists a bijection between these two.

Using that definition I can easily state that $\mathbb{N} \equiv \{ 2, 4, 6, \dotsc \} \subset \mathbb{N}$. Hence it can be seen that the bijection-based of equivalence is not that precise and sensitive: I just lost a countable infinity of numbers somewhere and yet it states the two sets are "almost the same".

Are there more precise, more sensitive notions of equivalence?

Xander Henderson
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Zazaeil
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  • Two sets A and B have the same cardinality if there exists a bijection (a.k.a., one-to-one correspondence) from A to B, that is, a function from A to B that is both injective and surjective. Such sets are said to be equipotent, equipollent, or equinumerous. What do you mean with "a more reliable notion of equivalence"? – Mauro ALLEGRANZA Oct 11 '21 at 12:57
  • I wouldn't use the expression "Common sense". I would say that it's a natural extension from finite to infinite sets. – Jean Marie Oct 11 '21 at 12:57
  • @MauroALLEGRANZA I mean such a notion that does not prove two sets are equivalent when they differ in an infinite number of elements. – Zazaeil Oct 11 '21 at 12:59
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    We do not call them "equivalent": we call them equinumerous and this does not mean that they are "equal". – Mauro ALLEGRANZA Oct 11 '21 at 13:00
  • ${ 1,2,3 }$ and ${ 4,5,6 }$ are equinuomerous also in the usual intuitive sense and they are not "equal". – Mauro ALLEGRANZA Oct 11 '21 at 13:01
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    There is no notion of "equivalence" of sets per se. There is the notion of having the size/cardinality which is the notion you are talking about. – Henno Brandsma Oct 11 '21 at 13:01
  • @MauroALLEGRANZA OK, just replace "equivalent" with the "equinumerous". They still differ in the infinite number of elements and they still have proven to be equinumerous. Does a stricter version of "equinumerous" exist? – Zazaeil Oct 11 '21 at 13:03
  • Indeed, one characteristic of an infinite set is that it can be put in one-to-one correspondence with a proper subset. – Matthew Leingang Oct 11 '21 at 13:03
  • @MatthewLeingang so it comes somewhere from the ZFC axiomatization? – Zazaeil Oct 11 '21 at 13:04
  • No, it's a question of definition not axioms. And no-one has proposed a theory based on another "equivalence". Just consensus. Deal with it. – Henno Brandsma Oct 11 '21 at 13:05
  • @HennoBrandsma so ZFC defines an (countably or uncountably) infite set as one that can be put in one-to-one correspondence with a proper subset? – Zazaeil Oct 11 '21 at 13:06
  • ZFC does not define anything. You define something in the framework of ZFC maybe, but this notion predates ZFC and is due to Cantor himself, the "founding father" of set theory. – Henno Brandsma Oct 11 '21 at 13:06
  • @HennoBrandsma I don't get it. Can you please be a little more clear? – Zazaeil Oct 11 '21 at 13:07
  • I think you misunderstand what ZFC is. – Henno Brandsma Oct 11 '21 at 13:07
  • See H.Enderton, Elements of Set theory, page 129, Definition of equinumerous. – Mauro ALLEGRANZA Oct 11 '21 at 13:12

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You could define two sets to be equivalent if their symmetric difference is finite. This does differentiate between $\mathbb{N}$ and $2\mathbb{N}$; however, it does not differentiate between finite sets. If that does not suit you, you could define two sets to be equivalent if they can be put in a bijection and their symmetric difference is finite, which gives the same equivalence classes for infinite sets, plus the usual finite sets.

However, I do not think that this definition leads to any interesting results - in a sense, it is almost the same as set equality. Indeed, it seems to me that one of the reasons why the usual definition of cardinality leads to such a fascinating theory is because of how coarse it is, while not being so coarse as to be useless.

Dániel G.
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  • [+1] I like very much the philosophy of the second paragraph. It is exactly the same type of answer one could give to the question in boolean algebra"why $p \implies q$ is the same as $\lnot p \lor q$" (particular case "why is False implying anything" ?) – Jean Marie Oct 11 '21 at 13:24