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While reading I came across in a book with two different definitions about the same mathematical object. This kinda make feel anxious. Let me explain precisely what I mean. Cartesian product can be defined in two different but I assume equivalent ways.

For the two sets case AxB:

1.- The set of all ordered pairs such that x belongs to A and y to B.

2.- The set of all functions f with domain {1,2} and image R.

The two sets are isomorphic in terms of sets and we can regard them as the same thing. But it's this notion of "sameness" that let me wondering about. Are these two definitions actually equals or perhaps what it's the same is the object they describe. In other words, can we show that two definitions are equal without showing necessarily that the two objects they describe are equal?.

Eric Wofsey
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Brian de León
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    Isomorphy is what you truly should consider as "equality". Set theoretic equality is close to meaningless when it comes to mathematical ideas. Math is about how objects behave, not about what objects are. So objects which behave exactly the same way should be treated as essentially equal. – Vercassivelaunos Nov 05 '20 at 19:20
  • Thanks for answering. It makes sense then. The objects are not equal but as you say this isomorphism is what matters. Then when dealing with this, one should be careful when saying they are the same we should say they are isomorphic but not the same thing. – Brian de León Nov 05 '20 at 19:33
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    But you have to be precise about "isomorphic". For example, $\mathbb{N}\times\mathbb{N}$ is isomorphic to $\mathbb{N}$ in the category of sets. – Stefan Nov 05 '20 at 19:45
  • What is $R$ in your question? – Christian Blatter Nov 05 '20 at 19:46
  • @ChristianBlatter The set of reals Mr. Blatter – Brian de León Nov 05 '20 at 20:01
  • @Stefan be precise about isomorphic? Sorry I don't get it right away – Brian de León Nov 05 '20 at 20:02
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    Be precise about what you mean by isomorphic. You can't say that every definition that produces a set isomorphic to $A\times B$ is a definition of the cartesian product of $A$ and $B$, because isomorphic in terms of sets just means having the same number of elements. – Stefan Nov 05 '20 at 21:04
  • Yes @Stefan. Isomorphic in terms of sets. Sorry for not being clear, I'll edit the question. My question is more about an absolute sameness. – Brian de León Nov 05 '20 at 21:57
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    In my mind, I would just use definition 1. Sometimes I am bothered when mathematicians say "we will not distinguish between" two different but closely related mathematical objects. What does it mean to "not distinguish" them? I can't just pretend they are actually the same because I know they are not. Here is how I interpret that phrase: we sometimes use the same symbol or the same name for two different but related objects, and we hope that context makes it clear which object is being referred to. e.g., the sets defined in these two definitions might both be given the same name $A \times B$. – littleO Nov 06 '20 at 05:45
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    These are not equivalent as written. Your second definition should read codomain $\mathbb{R}$ (since it is impossible for a function with a finite domain to have range $\mathbb{R}$); and then it provides a definition for the particular Cartesian product $\mathbb{R} \times \mathbb{R}$. – xxxxxxxxx Nov 06 '20 at 05:55
  • @Brian: What I'm trying to say is that "isomorphic in terms of sets" is not enough, or rather that the notion of "cartesian product" consists of more than just a set. For example, in the category-theoretic sense, it consists of a set and two mappings (the projections). In your example, these mappings would be $(a,b)\mapsto a$ and $(a,b)\mapsto b$ in 1., and $f\mapsto f(1)$ and $f\mapsto f(2)$ in 2.. The two definitions being equivalent does not follow from the existence of an isomorphism of the sets alone, but from the existence of an isomorphism of the sets which commutes with the projections – Stefan Nov 06 '20 at 23:06

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