Nonisomorphism of real numbers and reals without 0 (the only symbol is the < symbol)












4












$begingroup$


Why are $(mathbb{R}, <)$ and $({ x in mathbb{R} | x neq 0}, <)$ not isomorphic?



My approach is to assume that there is an isomorphism and attempt to derive a contradiction by showing that there is a formula which the first structure satisfies but the second does not. I'm having trouble finding one.






logic first-order-logic order-theory model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isomorphism as what? Ordered sets? Groups?
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:24












  • $begingroup$
    Sorry, as structures in a first order language.
    $endgroup$
    – user514088
    Nov 30 '18 at 12:25










  • $begingroup$
    I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:26












  • $begingroup$
    The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:27
















4












$begingroup$


Why are $(mathbb{R}, <)$ and $({ x in mathbb{R} | x neq 0}, <)$ not isomorphic?



My approach is to assume that there is an isomorphism and attempt to derive a contradiction by showing that there is a formula which the first structure satisfies but the second does not. I'm having trouble finding one.






logic first-order-logic order-theory model-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Isomorphism as what? Ordered sets? Groups?
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:24












  • $begingroup$
    Sorry, as structures in a first order language.
    $endgroup$
    – user514088
    Nov 30 '18 at 12:25










  • $begingroup$
    I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:26












  • $begingroup$
    The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:27














4












4








4





$begingroup$


Why are $(mathbb{R}, <)$ and $({ x in mathbb{R} | x neq 0}, <)$ not isomorphic?



My approach is to assume that there is an isomorphism and attempt to derive a contradiction by showing that there is a formula which the first structure satisfies but the second does not. I'm having trouble finding one.






logic first-order-logic order-theory model-theory







share|cite|improve this question











$endgroup$




Why are $(mathbb{R}, <)$ and $({ x in mathbb{R} | x neq 0}, <)$ not isomorphic?



My approach is to assume that there is an isomorphism and attempt to derive a contradiction by showing that there is a formula which the first structure satisfies but the second does not. I'm having trouble finding one.






logic first-order-logic order-theory model-theory




logic first-order-logic order-theory model-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 13:25









Scientifica

6,37641335




6,37641335










asked Nov 30 '18 at 12:19









user514088user514088

211




211












  • $begingroup$
    Isomorphism as what? Ordered sets? Groups?
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:24












  • $begingroup$
    Sorry, as structures in a first order language.
    $endgroup$
    – user514088
    Nov 30 '18 at 12:25










  • $begingroup$
    I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:26












  • $begingroup$
    The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:27


















  • $begingroup$
    Isomorphism as what? Ordered sets? Groups?
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:24












  • $begingroup$
    Sorry, as structures in a first order language.
    $endgroup$
    – user514088
    Nov 30 '18 at 12:25










  • $begingroup$
    I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
    $endgroup$
    – Arthur
    Nov 30 '18 at 12:26












  • $begingroup$
    The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
    $endgroup$
    – Leo163
    Nov 30 '18 at 12:27
















$begingroup$
Isomorphism as what? Ordered sets? Groups?
$endgroup$
– Arthur
Nov 30 '18 at 12:24






$begingroup$
Isomorphism as what? Ordered sets? Groups?
$endgroup$
– Arthur
Nov 30 '18 at 12:24














$begingroup$
Sorry, as structures in a first order language.
$endgroup$
– user514088
Nov 30 '18 at 12:25




$begingroup$
Sorry, as structures in a first order language.
$endgroup$
– user514088
Nov 30 '18 at 12:25












$begingroup$
I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
$endgroup$
– Arthur
Nov 30 '18 at 12:26






$begingroup$
I'm a bit fuzzy on what exactly kind of thing you're looking for, but would the least upper bound property be useful? The latter of the two has non-empty, upwards bounded subsets with no supremum.
$endgroup$
– Arthur
Nov 30 '18 at 12:26














$begingroup$
The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
$endgroup$
– Leo163
Nov 30 '18 at 12:27




$begingroup$
The proof is basically the same as if you had to prove that they are not homeomorphic (after all, the topology on $mathbb{R}$ is induced by the order)
$endgroup$
– Leo163
Nov 30 '18 at 12:27










1 Answer
1






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oldest

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5












$begingroup$

Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.



So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(mathbb Q,<)$ and $(mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $mathbb Q$ is countable but not $mathbb R$. You can't say in the first order logic with the language ${<}$ that a model is countable or not.



Same thing goes for $(mathbb R,<)$ and $(mathbb Rbackslash{0}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.



Leo163's comment gives another proof. The usual topology on $mathbb R$ coincides with its order topology, and same goes for $mathbb Rbackslash{0}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ were isomorphic, $mathbb R$ would be homeomorphic to $mathbb Rbackslash{0}$, but one is connected and the other is not.






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    1 Answer
    1






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    active

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    active

    oldest

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    5












    $begingroup$

    Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.



    So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(mathbb Q,<)$ and $(mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $mathbb Q$ is countable but not $mathbb R$. You can't say in the first order logic with the language ${<}$ that a model is countable or not.



    Same thing goes for $(mathbb R,<)$ and $(mathbb Rbackslash{0}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.



    Leo163's comment gives another proof. The usual topology on $mathbb R$ coincides with its order topology, and same goes for $mathbb Rbackslash{0}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ were isomorphic, $mathbb R$ would be homeomorphic to $mathbb Rbackslash{0}$, but one is connected and the other is not.






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.



      So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(mathbb Q,<)$ and $(mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $mathbb Q$ is countable but not $mathbb R$. You can't say in the first order logic with the language ${<}$ that a model is countable or not.



      Same thing goes for $(mathbb R,<)$ and $(mathbb Rbackslash{0}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.



      Leo163's comment gives another proof. The usual topology on $mathbb R$ coincides with its order topology, and same goes for $mathbb Rbackslash{0}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ were isomorphic, $mathbb R$ would be homeomorphic to $mathbb Rbackslash{0}$, but one is connected and the other is not.






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.



        So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(mathbb Q,<)$ and $(mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $mathbb Q$ is countable but not $mathbb R$. You can't say in the first order logic with the language ${<}$ that a model is countable or not.



        Same thing goes for $(mathbb R,<)$ and $(mathbb Rbackslash{0}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.



        Leo163's comment gives another proof. The usual topology on $mathbb R$ coincides with its order topology, and same goes for $mathbb Rbackslash{0}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ were isomorphic, $mathbb R$ would be homeomorphic to $mathbb Rbackslash{0}$, but one is connected and the other is not.






        share|cite|improve this answer









        $endgroup$



        Your idea is good. Indeed, two isomorphic structures satisfy the same formulas. But pay attention, the converse is not true. For example, if you covered the notion of back-and-forth, you may have shown that the theory of dense linear orders without extremities is complete. Both $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ are such orders, so they are elementarily equivalent, and hence satisfy the same formulas.



        So how can you show that they are not isomorphic? By showing that they don't satisfy the same properties that are not necessarily expressed in first order logic. For example, for the same reasons as above, $(mathbb Q,<)$ and $(mathbb R,<)$ are both elementarily equivalent, yet they're clearly not isomorphic because $mathbb Q$ is countable but not $mathbb R$. You can't say in the first order logic with the language ${<}$ that a model is countable or not.



        Same thing goes for $(mathbb R,<)$ and $(mathbb Rbackslash{0}$,<). One of them satisfies the least upper bound property, but not the other. This shows that they are not isomorphic.



        Leo163's comment gives another proof. The usual topology on $mathbb R$ coincides with its order topology, and same goes for $mathbb Rbackslash{0}$. The way this topology is defined, you can see that if two totally ordered sets are isomorphic (as ordered sets) then they're homeomorphic (as topological spaces, where the topology of each one is the order topology). So, if $(mathbb R,<)$ and $(mathbb Rbackslash{0},<)$ were isomorphic, $mathbb R$ would be homeomorphic to $mathbb Rbackslash{0}$, but one is connected and the other is not.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 13:25









        ScientificaScientifica

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        6,37641335






























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