Does the set $\mathbb{Z}$ satisfy the completeness axiom? The Completeness Axiom states:

A set $\mathcal{A}$ satisfies the Completeness Axiom if for every of its non-trivial and bounded subsets, a supremum exists in $\mathcal{A}$. 

If this is the statement for the completeness axiom then doesn't this imply that the set $\mathbb{Z}$ also satisfies the axiom? (Because one can easily find a supremum of a non-trivial subset of $\mathbb{Z}$.)
But then I read somewhere that $\mathbb{R}$ is the only field that is complete and if there exists another field satisfying the axiom, then it is isomorphic to the field $\mathbb{R}$ . I can't seem to understand this clearly. Does this mean that $\mathbb{R} \cong \mathbb{Z}$? Where am I going wrong?
 A: $\mathbb{Z}$ does satisfy the completeness axiom. 
However its elements do not have multiplicative inverses (except $1$ and $-1$) and so it is not a field. 
A: Yes, the ordered set $(\mathbb Z,\le)$ is indeed a complete order. Every bounded subset has a supremum and infimum.
Note that also $(\mathbb N,\le)$ has the same property. There is a least and last if a set is bounded.
However nor $\mathbb N$ neither $\mathbb Z$ are fields. The number $2$ is in both, but $\dfrac12$ is in neither.

The completeness axiom is about orders. There are many orders which are complete. They do not even have to be linear, they can be partial ordered sets just as well (note that $P(\mathbb N)$ as a power set is complete under $\subseteq$ in a very similar fashion).
The fact to which you refer is that $\mathbb R$ is the unique (up to isomorphism) ordered field which is order-complete. Therefore any non-isomorphic order which is complete cannot be the order of a field (i.e. we cannot define addition and multiplication in a way which plays nice with the order)
