Let $X$ be the set of all $A\subset\mathbb Z$ that are bounded above; that is, $A\in X$ iff $A\subset\mathbb Z$ and $\exists\max A$. Define the following operation, as the sum of two such sets: $A\oplus B=(A\triangle B)\oplus s(A\cap B)$, where the function $s$ sends all the elements of a set, into the respective succesors. Explicitly, if $A=\{a\}_{\{a\in A\}}$, then $sA=\{a+1\}_{\{a\in A\}}$. The trick to this operation is that it ends in a finite number of recursive steps, given $A,B$ are finite. This means that when we apply the definition $A\oplus B$ we get another addition of sets $A´\oplus B´$. It can be proven, however, that after a finite number of itereations the term $B^{(n)}$ is the empty set. Leaving us as final answer the set $A^{(n)}$.

This operation makes $X$ order and operation isomorphic to the set of positive real numbers. The order is defined in terms of the symmetric difference, and can be seen in the following link, where an explicit construction of the supremum is described:

Describe an infinite process of denumerable steps, that defines supremum property for the following set X (supremum of any subset of X exists)

The isomorphism is quite natural, since every positive real number is expressable as a sum of integer powers of 2, we map the real number to the set of integers that are the powers in its expansion of powers of 2. Therefore, natural numbers are bounded subsets of $\mathbb N$, while numbers in the continuum $[0,1]$ are arbitrary subsets of $-\mathbb N$.

  • $\begingroup$ Your question is far from clear. If by "operation isomorphic to the set of positive real numbers" you mean that you already know that the operation is isomorphic to real number addition, then there is nothing to prove. Please provide more context. $\endgroup$ – Rob Arthan May 29 '18 at 19:57
  • $\begingroup$ You are right. I have proven the isomorphism, through other means but would like to see a direct proof of the associativity property. In other words one can see that the structure is indeed isomoprhic to the real number structure, only that to prove associativity directly , seems to have only one solution for proving it, and I would like to see if any one else can see a different solution. I have read your construction on Eudoxus Reals. Great! $\endgroup$ – Juan Ramirez May 29 '18 at 20:04
  • $\begingroup$ It is mentioned in your discussion of termination after finitely many steps that when the second argument is the empty set, the result is defined by the first argument. However this should be stated explicitly in the definition of $\oplus$, because it doesn't follow from the recursive branch of the definition. $\endgroup$ – hardmath May 29 '18 at 20:10
  • $\begingroup$ @JuanRamirez: thanks for the kind words about my paper. MSE is sometimes not very kind to questions about personal research, so I suggest you edit your question to say a bit more about the context and about the isomorphism. $\endgroup$ – Rob Arthan May 29 '18 at 20:15
  • $\begingroup$ Thank you, ill check my question and keep it in mind. $\endgroup$ – Juan Ramirez May 29 '18 at 20:16

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