Can we interpret sets of the same cardinality as distinct representations of the same set?

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Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,cdot)$ is isomorphic to $(B,oplus,odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:



Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $varphi:Ato B$ between them, because, just as the apparently distinct sets $1,, 2,, 3,, cdots$ and $I,, II,, III,, cdots$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.



One can argue that this interpretation is wrong since there is a bijection between $mathbbN$ and $mathbbQ$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $mathbbN$ and $mathbbQ$ are not simply sets, but an algebraic structures. For example, $mathbbN$ is actually a triple $(mathbbN,+,cdot)$ in which $mathbbN=1,, 2,,3,, cdots$ and $+$ and $cdot$ are binary operations defined in $mathbbN$ with which we can build a well-ordering $leq $ order in $mathbbN$. So, as algebraic structures, $mathbbN$ and $mathbbQ$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:mathbbNtomathbbQ$ I can represent all elements of $mathbbQ$ with the symbols $1,, 2,, 3,, cdots$. To do this, I can simply represent an element $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ that satisfies $f(n)=q$.



If we think well it is not possible to infer that $mathbbN=1,, 2,,3,, cdots$ from the axioms of Peano. What happens is that we use the symbols $1,, 2,, 3,, cdots$ to represent the natural numbers which is the same attitude as representing $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ satisfying $f(n)=q$. Therefore, if we consider $mathbbN$ and $mathbbQ$ only as sets (and not as algebraic structures), we can say that $mathbbN=mathbbQ$ since it is possible to represent $mathbbQ$ with the same symbols used to represent the elements of $mathbbN$.



My question is: can we interpret sets of the same cardinality as distinct representations of the same set? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,cdot)$ and $(L,odot)$ are isomorphic, then $(G,cdot)$ and $(L,odot)$ are distinct representations of the same group.



EDIT: I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.







share|cite|improve this question


















  • 1




    Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
    – Andrés E. Caicedo
    Aug 10 at 20:21






  • 1




    You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
    – zzuussee
    Aug 10 at 21:13










  • @zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
    – rfloc
    Aug 10 at 21:23






  • 1




    Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
    – zzuussee
    Aug 10 at 21:28






  • 1




    @mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
    – rfloc
    Aug 11 at 2:55














up vote
1
down vote

favorite












Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,cdot)$ is isomorphic to $(B,oplus,odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:



Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $varphi:Ato B$ between them, because, just as the apparently distinct sets $1,, 2,, 3,, cdots$ and $I,, II,, III,, cdots$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.



One can argue that this interpretation is wrong since there is a bijection between $mathbbN$ and $mathbbQ$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $mathbbN$ and $mathbbQ$ are not simply sets, but an algebraic structures. For example, $mathbbN$ is actually a triple $(mathbbN,+,cdot)$ in which $mathbbN=1,, 2,,3,, cdots$ and $+$ and $cdot$ are binary operations defined in $mathbbN$ with which we can build a well-ordering $leq $ order in $mathbbN$. So, as algebraic structures, $mathbbN$ and $mathbbQ$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:mathbbNtomathbbQ$ I can represent all elements of $mathbbQ$ with the symbols $1,, 2,, 3,, cdots$. To do this, I can simply represent an element $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ that satisfies $f(n)=q$.



If we think well it is not possible to infer that $mathbbN=1,, 2,,3,, cdots$ from the axioms of Peano. What happens is that we use the symbols $1,, 2,, 3,, cdots$ to represent the natural numbers which is the same attitude as representing $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ satisfying $f(n)=q$. Therefore, if we consider $mathbbN$ and $mathbbQ$ only as sets (and not as algebraic structures), we can say that $mathbbN=mathbbQ$ since it is possible to represent $mathbbQ$ with the same symbols used to represent the elements of $mathbbN$.



My question is: can we interpret sets of the same cardinality as distinct representations of the same set? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,cdot)$ and $(L,odot)$ are isomorphic, then $(G,cdot)$ and $(L,odot)$ are distinct representations of the same group.



EDIT: I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.







share|cite|improve this question


















  • 1




    Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
    – Andrés E. Caicedo
    Aug 10 at 20:21






  • 1




    You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
    – zzuussee
    Aug 10 at 21:13










  • @zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
    – rfloc
    Aug 10 at 21:23






  • 1




    Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
    – zzuussee
    Aug 10 at 21:28






  • 1




    @mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
    – rfloc
    Aug 11 at 2:55












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,cdot)$ is isomorphic to $(B,oplus,odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:



Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $varphi:Ato B$ between them, because, just as the apparently distinct sets $1,, 2,, 3,, cdots$ and $I,, II,, III,, cdots$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.



One can argue that this interpretation is wrong since there is a bijection between $mathbbN$ and $mathbbQ$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $mathbbN$ and $mathbbQ$ are not simply sets, but an algebraic structures. For example, $mathbbN$ is actually a triple $(mathbbN,+,cdot)$ in which $mathbbN=1,, 2,,3,, cdots$ and $+$ and $cdot$ are binary operations defined in $mathbbN$ with which we can build a well-ordering $leq $ order in $mathbbN$. So, as algebraic structures, $mathbbN$ and $mathbbQ$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:mathbbNtomathbbQ$ I can represent all elements of $mathbbQ$ with the symbols $1,, 2,, 3,, cdots$. To do this, I can simply represent an element $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ that satisfies $f(n)=q$.



If we think well it is not possible to infer that $mathbbN=1,, 2,,3,, cdots$ from the axioms of Peano. What happens is that we use the symbols $1,, 2,, 3,, cdots$ to represent the natural numbers which is the same attitude as representing $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ satisfying $f(n)=q$. Therefore, if we consider $mathbbN$ and $mathbbQ$ only as sets (and not as algebraic structures), we can say that $mathbbN=mathbbQ$ since it is possible to represent $mathbbQ$ with the same symbols used to represent the elements of $mathbbN$.



My question is: can we interpret sets of the same cardinality as distinct representations of the same set? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,cdot)$ and $(L,odot)$ are isomorphic, then $(G,cdot)$ and $(L,odot)$ are distinct representations of the same group.



EDIT: I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.







share|cite|improve this question














Usually mathematicians consider isomorphic fields as equal fields. That is, if the $(A,+,cdot)$ is isomorphic to $(B,oplus,odot)$, then I can consider those fields as equals. Thinking about it, I thought about the following interpretation:



Let $A$ and $B$ be two sets. I think we can interpret that $A$ and $B$, considered only as sets, are the same set if there is a bijection $varphi:Ato B$ between them, because, just as the apparently distinct sets $1,, 2,, 3,, cdots$ and $I,, II,, III,, cdots$ can represent the set of natural numbers, $A$ and $B$, if there is a bijection between them, can be understood as distinct representations of the same set.



One can argue that this interpretation is wrong since there is a bijection between $mathbbN$ and $mathbbQ$ and these sets are clearly distinct since the first one has the well-ordering principle the second one doesn't. However $mathbbN$ and $mathbbQ$ are not simply sets, but an algebraic structures. For example, $mathbbN$ is actually a triple $(mathbbN,+,cdot)$ in which $mathbbN=1,, 2,,3,, cdots$ and $+$ and $cdot$ are binary operations defined in $mathbbN$ with which we can build a well-ordering $leq $ order in $mathbbN$. So, as algebraic structures, $mathbbN$ and $mathbbQ$ are indeed distinct, but as simply sets we can consider them the same, because since there is an bijection $f:mathbbNtomathbbQ$ I can represent all elements of $mathbbQ$ with the symbols $1,, 2,, 3,, cdots$. To do this, I can simply represent an element $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ that satisfies $f(n)=q$.



If we think well it is not possible to infer that $mathbbN=1,, 2,,3,, cdots$ from the axioms of Peano. What happens is that we use the symbols $1,, 2,, 3,, cdots$ to represent the natural numbers which is the same attitude as representing $qinmathbbQ$ by an $nin 1,, 2,, 3,, cdots$ satisfying $f(n)=q$. Therefore, if we consider $mathbbN$ and $mathbbQ$ only as sets (and not as algebraic structures), we can say that $mathbbN=mathbbQ$ since it is possible to represent $mathbbQ$ with the same symbols used to represent the elements of $mathbbN$.



My question is: can we interpret sets of the same cardinality as distinct representations of the same set? Because if the answer is affirmative then it becomes easier to understand why we can consider isomorphic fields or isomorphic groups as equals. Thinking about groups I think it is also possible to think that if the groups $(G,cdot)$ and $(L,odot)$ are isomorphic, then $(G,cdot)$ and $(L,odot)$ are distinct representations of the same group.



EDIT: I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 14 at 3:41









BCLC

6,77822073




6,77822073










asked Aug 10 at 20:14









rfloc

1238




1238







  • 1




    Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
    – Andrés E. Caicedo
    Aug 10 at 20:21






  • 1




    You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
    – zzuussee
    Aug 10 at 21:13










  • @zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
    – rfloc
    Aug 10 at 21:23






  • 1




    Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
    – zzuussee
    Aug 10 at 21:28






  • 1




    @mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
    – rfloc
    Aug 11 at 2:55












  • 1




    Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
    – Andrés E. Caicedo
    Aug 10 at 20:21






  • 1




    You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
    – zzuussee
    Aug 10 at 21:13










  • @zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
    – rfloc
    Aug 10 at 21:23






  • 1




    Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
    – zzuussee
    Aug 10 at 21:28






  • 1




    @mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
    – rfloc
    Aug 11 at 2:55







1




1




Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
– Andrés E. Caicedo
Aug 10 at 20:21




Your question is not really about set theory (where we very much consider different sets as being distinct), but there may be a way of saying something interesting from a category-theoretic point of view.
– Andrés E. Caicedo
Aug 10 at 20:21




1




1




You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
– zzuussee
Aug 10 at 21:13




You really don't consider isomorphic structures "equal", you consider them as really just isomorphic, that is in some sense a change of notation. But they may be very much not the same objects, set theoretically.
– zzuussee
Aug 10 at 21:13












@zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
– rfloc
Aug 10 at 21:23




@zzuussee "that is in some sense a change of notation". It is precisely for this reason that it makes sense for me to consider two isomorphic groups as distinct representations of the same group.
– rfloc
Aug 10 at 21:23




1




1




Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
– zzuussee
Aug 10 at 21:28




Yes, on a meta-level they can be regarded as the same object, which is basically the whole idea of isomorphy. But, regarding their actual encoding in some set theory, they are very much not the same. I agree (partially) with your assertions, I just was not comfortable with equal (in a strict sense).
– zzuussee
Aug 10 at 21:28




1




1




@mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
– rfloc
Aug 11 at 2:55




@mr_e_man All I'm talking about is an attempt to understand the attitude of various mathematicians in treating different objects as if they were the same just because there is an isomorphism between them. For this reason I said "Can we INTERPRET...". A ubiquitous example in books dealing with fields is the statement of $Fsubseteq L$ when in fact there is a monomorphism rather than a inclusion of the field $F$ in $L$. These sets you cited made me realize that it makes more sense to use this interpretation in more complex structures rather than using it in sets.
– rfloc
Aug 11 at 2:55










3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










The thing to recognize here is that the mathematical notion of equality has bifuricated.



In addition to the traditional notion of equality, we now recognize the value in considering the distinct notion of having an equivalence between two things.



We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification.



For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.



If you only ever ask questions about equivalence, never about equality, then you will indeed see $mathbbN$ and $mathbbQ$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what you mean by "$mathbbN$ is the same as $mathbbQ$" is a bijection $mathbbN to mathbbQ$.



Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.



However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.






share|cite|improve this answer





























    up vote
    2
    down vote













    Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.



    Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.



    You have answered your question when comparing the two sets , natural numbers and rational numbers.



    They are not the same set.






    share|cite|improve this answer




















    • As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
      – rfloc
      Aug 10 at 23:33

















    up vote
    2
    down vote













    I suppose you know the concept of a category. If not, I strongly recommend to consult any book on this thematic area. Your question suggests that you regard objects of a category $mathcalC$ as essentially equal if they are isomorphic in $mathcalC$.



    The category $Set$ has as objects all sets and as morphisms all functions between sets. The isomorphisms in $Set$ are precisely the bijections. This means that two sets are isomorphic in $Set$ if and only if they have the same cardinality. In that sense they are essentially equal. Your example $mathbbN$ and $mathbbQ$ does not concern the category $Set$, but the category $OrdSet$ of ordered sets (morphisms are order preserving functions). In this category $mathbbN$ and $mathbbQ$ with their natural orders are not isomorphic.






    share|cite|improve this answer






















    • I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
      – rfloc
      Aug 11 at 1:59










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote



    accepted










    The thing to recognize here is that the mathematical notion of equality has bifuricated.



    In addition to the traditional notion of equality, we now recognize the value in considering the distinct notion of having an equivalence between two things.



    We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification.



    For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.



    If you only ever ask questions about equivalence, never about equality, then you will indeed see $mathbbN$ and $mathbbQ$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what you mean by "$mathbbN$ is the same as $mathbbQ$" is a bijection $mathbbN to mathbbQ$.



    Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.



    However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.






    share|cite|improve this answer


























      up vote
      2
      down vote



      accepted










      The thing to recognize here is that the mathematical notion of equality has bifuricated.



      In addition to the traditional notion of equality, we now recognize the value in considering the distinct notion of having an equivalence between two things.



      We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification.



      For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.



      If you only ever ask questions about equivalence, never about equality, then you will indeed see $mathbbN$ and $mathbbQ$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what you mean by "$mathbbN$ is the same as $mathbbQ$" is a bijection $mathbbN to mathbbQ$.



      Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.



      However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        The thing to recognize here is that the mathematical notion of equality has bifuricated.



        In addition to the traditional notion of equality, we now recognize the value in considering the distinct notion of having an equivalence between two things.



        We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification.



        For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.



        If you only ever ask questions about equivalence, never about equality, then you will indeed see $mathbbN$ and $mathbbQ$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what you mean by "$mathbbN$ is the same as $mathbbQ$" is a bijection $mathbbN to mathbbQ$.



        Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.



        However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.






        share|cite|improve this answer














        The thing to recognize here is that the mathematical notion of equality has bifuricated.



        In addition to the traditional notion of equality, we now recognize the value in considering the distinct notion of having an equivalence between two things.



        We might also consider the proposition asserting two things are equivalent (i.e. that there exists an equivalence between them), but in my opinion practice has shown that's not the right notion; it's merely a frequently useful simplification.



        For sets, the right notion of equivalence is a bijective function. More generally for algebraic structures (or objects of a category) the right notion is that of an isomorphism.



        If you only ever ask questions about equivalence, never about equality, then you will indeed see $mathbbN$ and $mathbbQ$ as being "the same" — and, in fact, see them as being "the same" in lots of different ways — because what you mean by "$mathbbN$ is the same as $mathbbQ$" is a bijection $mathbbN to mathbbQ$.



        Furthermore, you do see mathematicians in such settings repurpose the term $=$ to mean equivalence, and maybe even pronounce it as "equals" as well.



        However, if you find yourself needing to also consider the traditional notion of equality — or, at least, want to converse with mathematicians who use the traditional notion — then you should use "equivalence" for the notion of sameness that you have.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 11 at 3:01

























        answered Aug 11 at 2:50









        Hurkyl

        108k9112254




        108k9112254




















            up vote
            2
            down vote













            Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.



            Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.



            You have answered your question when comparing the two sets , natural numbers and rational numbers.



            They are not the same set.






            share|cite|improve this answer




















            • As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
              – rfloc
              Aug 10 at 23:33














            up vote
            2
            down vote













            Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.



            Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.



            You have answered your question when comparing the two sets , natural numbers and rational numbers.



            They are not the same set.






            share|cite|improve this answer




















            • As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
              – rfloc
              Aug 10 at 23:33












            up vote
            2
            down vote










            up vote
            2
            down vote









            Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.



            Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.



            You have answered your question when comparing the two sets , natural numbers and rational numbers.



            They are not the same set.






            share|cite|improve this answer












            Two sets are considered equal iff every element of the first set is an element of the second set and every element of the second set is an element of the first set.



            Sets of the same Cardinality are not the same sets but they enjoy having a one to one correspondence between them.



            You have answered your question when comparing the two sets , natural numbers and rational numbers.



            They are not the same set.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 10 at 22:01









            Mohammad Riazi-Kermani

            28.2k41852




            28.2k41852











            • As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
              – rfloc
              Aug 10 at 23:33
















            • As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
              – rfloc
              Aug 10 at 23:33















            As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
            – rfloc
            Aug 10 at 23:33




            As I said in my question, natural numbers and rational numbers as algebraic structures are obviously distinct. But treating them only as sets it is possible to interpret them as equals (at least it makes sense to me)
            – rfloc
            Aug 10 at 23:33










            up vote
            2
            down vote













            I suppose you know the concept of a category. If not, I strongly recommend to consult any book on this thematic area. Your question suggests that you regard objects of a category $mathcalC$ as essentially equal if they are isomorphic in $mathcalC$.



            The category $Set$ has as objects all sets and as morphisms all functions between sets. The isomorphisms in $Set$ are precisely the bijections. This means that two sets are isomorphic in $Set$ if and only if they have the same cardinality. In that sense they are essentially equal. Your example $mathbbN$ and $mathbbQ$ does not concern the category $Set$, but the category $OrdSet$ of ordered sets (morphisms are order preserving functions). In this category $mathbbN$ and $mathbbQ$ with their natural orders are not isomorphic.






            share|cite|improve this answer






















            • I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
              – rfloc
              Aug 11 at 1:59














            up vote
            2
            down vote













            I suppose you know the concept of a category. If not, I strongly recommend to consult any book on this thematic area. Your question suggests that you regard objects of a category $mathcalC$ as essentially equal if they are isomorphic in $mathcalC$.



            The category $Set$ has as objects all sets and as morphisms all functions between sets. The isomorphisms in $Set$ are precisely the bijections. This means that two sets are isomorphic in $Set$ if and only if they have the same cardinality. In that sense they are essentially equal. Your example $mathbbN$ and $mathbbQ$ does not concern the category $Set$, but the category $OrdSet$ of ordered sets (morphisms are order preserving functions). In this category $mathbbN$ and $mathbbQ$ with their natural orders are not isomorphic.






            share|cite|improve this answer






















            • I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
              – rfloc
              Aug 11 at 1:59












            up vote
            2
            down vote










            up vote
            2
            down vote









            I suppose you know the concept of a category. If not, I strongly recommend to consult any book on this thematic area. Your question suggests that you regard objects of a category $mathcalC$ as essentially equal if they are isomorphic in $mathcalC$.



            The category $Set$ has as objects all sets and as morphisms all functions between sets. The isomorphisms in $Set$ are precisely the bijections. This means that two sets are isomorphic in $Set$ if and only if they have the same cardinality. In that sense they are essentially equal. Your example $mathbbN$ and $mathbbQ$ does not concern the category $Set$, but the category $OrdSet$ of ordered sets (morphisms are order preserving functions). In this category $mathbbN$ and $mathbbQ$ with their natural orders are not isomorphic.






            share|cite|improve this answer














            I suppose you know the concept of a category. If not, I strongly recommend to consult any book on this thematic area. Your question suggests that you regard objects of a category $mathcalC$ as essentially equal if they are isomorphic in $mathcalC$.



            The category $Set$ has as objects all sets and as morphisms all functions between sets. The isomorphisms in $Set$ are precisely the bijections. This means that two sets are isomorphic in $Set$ if and only if they have the same cardinality. In that sense they are essentially equal. Your example $mathbbN$ and $mathbbQ$ does not concern the category $Set$, but the category $OrdSet$ of ordered sets (morphisms are order preserving functions). In this category $mathbbN$ and $mathbbQ$ with their natural orders are not isomorphic.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 11 at 14:14

























            answered Aug 10 at 22:25









            Paul Frost

            3,921421




            3,921421











            • I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
              – rfloc
              Aug 11 at 1:59
















            • I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
              – rfloc
              Aug 11 at 1:59















            I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
            – rfloc
            Aug 11 at 1:59




            I think I've found a way to express myself better. I can say that $mathbbQ=1,2,3,cdots $ in which the element $5$, for instance, is understood as the rational $q$ satisfying $q=f(5)$ ($f:mathbbNtomathbbQ$ is a bijection). Therefore, treating $mathbbQ$ and $mathbbN$ only as sets (and not as algebraic structures) I believe it makes sense to say $mathbbQ=mathbbN=1,2,3,cdots$.
            – rfloc
            Aug 11 at 1:59












             

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