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Joel David Hamkins
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If you meanintend literally to recover the addition operation of the given model $M$, then the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but a different addition functionsoperation $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals (or apply any permutation at all within a cluster, such as swapping two individuals only). That is, we define $M'$ so that $M$ will beis isomorphic to $M'$ by that shifting processthe map we considered. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

If you mean literally to recover the addition operation of the given model $M$, the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

If you intend literally to recover the addition operation of the given model $M$, then the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but a different addition operation $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals (or apply any permutation at all within a cluster, such as swapping two individuals only). That is, we define $M'$ so that $M$ is isomorphic to $M'$ by the map we considered. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

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Joel David Hamkins
  • 236.5k
  • 44
  • 777
  • 1.4k

If you mean literally to recover the addition operation of the given model $M$, the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

If you mean literally to recover the addition operation, the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

If you mean literally to recover the addition operation of the given model $M$, the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

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Joel David Hamkins
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TheIf you mean literally to recover the addition operation, the answer is nonegative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

The answer is no. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

If you mean literally to recover the addition operation, the answer is negative. For every nonstandard model of arithmetic $M$ there is another model $M'$ having exactly the same domain, the same clusters, with the same cluster order and cluster addition, but different addition functions $+^M\neq +^{M'}$.

Given any $M$, construct a model $M'$ by shifting the individuals within some nonstandard cluster, but fixing all other individuals. That is, $M$ will be isomorphic to $M'$ by that shifting process. So the models $M$ and $M'$ will have the same domain and the same clusters, and their clusters will be in the same order, and since cluster addition is well defined with respect to finite shifts, the models have the same cluster addition operation. But they have different addition operations.

Conclusion: the addition operation of the model is not determined by the order relation and addition operation on the clusters.

But perhaps you aim at a more subtle manner of recovering the structure, where you recover the addition of $M$ up to isomorphism, instead of the actual operation $x+y=z$ on individuals.

Source Link
Joel David Hamkins
  • 236.5k
  • 44
  • 777
  • 1.4k
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