Is it possible to construct order types using sets?

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It's possible to construct a "function" $f$ from the class of well-ordered sets to the class of all sets where for any well-ordered sets $A$ and $B$, we have $f(A)=f(B)$ if and only if $A$ and $B$ are order-isomorphic. (I put function in scare quotes because the domain and codomain are proper classes.) Von Neumann's construction of the ordinal numbers achieves this.



My question is, can we generalize this to the order types of all totally ordered sets? That is, can we construct a "function" $g$ from the class of totally ordered sets to the class of all sets where for any totally ordered sets $A$ and $B$, we have $g(A)=g(B)$ if and only if $A$ and $B$ are order-isomorphic? Of course such a function exists by the axiom of choice, but can we actually define such a function? And if so can we make it so that $g=f$ when restricted to the class of well-ordered sets?







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  • If you have global choice, sure.
    – Asaf Karagila♦
    Aug 22 at 12:13










  • @AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
    – Keshav Srinivasan
    Aug 22 at 13:06










  • If you have global choice, simply choose from the set of least rank or something like that.
    – Asaf Karagila♦
    Aug 22 at 13:50










  • @AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
    – Keshav Srinivasan
    Aug 22 at 13:52










  • Sure... Why not?
    – Asaf Karagila♦
    Aug 22 at 13:54














up vote
1
down vote

favorite
2












It's possible to construct a "function" $f$ from the class of well-ordered sets to the class of all sets where for any well-ordered sets $A$ and $B$, we have $f(A)=f(B)$ if and only if $A$ and $B$ are order-isomorphic. (I put function in scare quotes because the domain and codomain are proper classes.) Von Neumann's construction of the ordinal numbers achieves this.



My question is, can we generalize this to the order types of all totally ordered sets? That is, can we construct a "function" $g$ from the class of totally ordered sets to the class of all sets where for any totally ordered sets $A$ and $B$, we have $g(A)=g(B)$ if and only if $A$ and $B$ are order-isomorphic? Of course such a function exists by the axiom of choice, but can we actually define such a function? And if so can we make it so that $g=f$ when restricted to the class of well-ordered sets?







share|cite|improve this question




















  • If you have global choice, sure.
    – Asaf Karagila♦
    Aug 22 at 12:13










  • @AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
    – Keshav Srinivasan
    Aug 22 at 13:06










  • If you have global choice, simply choose from the set of least rank or something like that.
    – Asaf Karagila♦
    Aug 22 at 13:50










  • @AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
    – Keshav Srinivasan
    Aug 22 at 13:52










  • Sure... Why not?
    – Asaf Karagila♦
    Aug 22 at 13:54












up vote
1
down vote

favorite
2









up vote
1
down vote

favorite
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It's possible to construct a "function" $f$ from the class of well-ordered sets to the class of all sets where for any well-ordered sets $A$ and $B$, we have $f(A)=f(B)$ if and only if $A$ and $B$ are order-isomorphic. (I put function in scare quotes because the domain and codomain are proper classes.) Von Neumann's construction of the ordinal numbers achieves this.



My question is, can we generalize this to the order types of all totally ordered sets? That is, can we construct a "function" $g$ from the class of totally ordered sets to the class of all sets where for any totally ordered sets $A$ and $B$, we have $g(A)=g(B)$ if and only if $A$ and $B$ are order-isomorphic? Of course such a function exists by the axiom of choice, but can we actually define such a function? And if so can we make it so that $g=f$ when restricted to the class of well-ordered sets?







share|cite|improve this question












It's possible to construct a "function" $f$ from the class of well-ordered sets to the class of all sets where for any well-ordered sets $A$ and $B$, we have $f(A)=f(B)$ if and only if $A$ and $B$ are order-isomorphic. (I put function in scare quotes because the domain and codomain are proper classes.) Von Neumann's construction of the ordinal numbers achieves this.



My question is, can we generalize this to the order types of all totally ordered sets? That is, can we construct a "function" $g$ from the class of totally ordered sets to the class of all sets where for any totally ordered sets $A$ and $B$, we have $g(A)=g(B)$ if and only if $A$ and $B$ are order-isomorphic? Of course such a function exists by the axiom of choice, but can we actually define such a function? And if so can we make it so that $g=f$ when restricted to the class of well-ordered sets?









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 22 at 6:01









Keshav Srinivasan

1,84511339




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  • If you have global choice, sure.
    – Asaf Karagila♦
    Aug 22 at 12:13










  • @AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
    – Keshav Srinivasan
    Aug 22 at 13:06










  • If you have global choice, simply choose from the set of least rank or something like that.
    – Asaf Karagila♦
    Aug 22 at 13:50










  • @AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
    – Keshav Srinivasan
    Aug 22 at 13:52










  • Sure... Why not?
    – Asaf Karagila♦
    Aug 22 at 13:54
















  • If you have global choice, sure.
    – Asaf Karagila♦
    Aug 22 at 12:13










  • @AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
    – Keshav Srinivasan
    Aug 22 at 13:06










  • If you have global choice, simply choose from the set of least rank or something like that.
    – Asaf Karagila♦
    Aug 22 at 13:50










  • @AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
    – Keshav Srinivasan
    Aug 22 at 13:52










  • Sure... Why not?
    – Asaf Karagila♦
    Aug 22 at 13:54















If you have global choice, sure.
– Asaf Karagila♦
Aug 22 at 12:13




If you have global choice, sure.
– Asaf Karagila♦
Aug 22 at 12:13












@AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
– Keshav Srinivasan
Aug 22 at 13:06




@AsafKaragila But I don't just want to prove the existence of such a function, I want to explicitly construct such a function, akin to von Neumann's construction of the ordinals.
– Keshav Srinivasan
Aug 22 at 13:06












If you have global choice, simply choose from the set of least rank or something like that.
– Asaf Karagila♦
Aug 22 at 13:50




If you have global choice, simply choose from the set of least rank or something like that.
– Asaf Karagila♦
Aug 22 at 13:50












@AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
– Keshav Srinivasan
Aug 22 at 13:52




@AsafKaragila And can you make it so that $g$ restricted to well-ordered sets is equal to $f$?
– Keshav Srinivasan
Aug 22 at 13:52












Sure... Why not?
– Asaf Karagila♦
Aug 22 at 13:54




Sure... Why not?
– Asaf Karagila♦
Aug 22 at 13:54















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