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:
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 $-\mathbb N$.