# 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?

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You have to stop writing at the bottom of every post that you're a high school kid, etc. The fact that you're from Nepal is also irrelevant. – Asaf Karagila Jan 11 '12 at 22:00
I never meant that you're bragging (why would anyone brag of his origin nowadays anyway? :-)) It's just very irrelevant to the post. People will answer what and how they feel like. If you wish to receive elaborations to answers, just ask follow up questions in the comments to the answers. – Asaf Karagila Jan 11 '12 at 22:14
@DashDart It is very useful to say something about your level of knowledge, so that answerers can use that to decide how to best reply. But please put such general info in your profile, not in every question. And put it at the beginning of your profile so that answerers may quickly locate it. If specific remarks beyond your profile are needed for some question, then it is ok to elaborate in the question. But there is no need to repeat the general background info in your profile. – Bill Dubuque Jan 11 '12 at 22:43

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)

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I want to check my understanding. Is $\mathbb{Z}_p$ for prime $p$ complete? – tomcuchta Jan 12 '12 at 0:45
@tomcuchta: Do you mean $\{0,\ldots,p-1\}$ or the $p$-adic integers? Either way, you need to specify the order too. Both the interpretation yield a non-ordered ring (or group). – Asaf Karagila Jan 12 '12 at 11:04
$\{0,...,p-1\}$ with the order $0 < 1 < ... < p-1$. I suppose this does not work as an ordering because $(p-2)+3 = 1$ -- combining two larger numbers yields a smaller number? – tomcuchta Jan 12 '12 at 11:10
@tomcuchta: Simply as an ordered set (without the addition) this is indeed a complete order. Every finite linear order is complete. However this order does not respect the addition, since $1<1+1<1+1+1<1+\ldots+1=0<1$ (for $p$ many times). – Asaf Karagila Jan 12 '12 at 11:50

$\mathbb{Z}$ does satisfy the completeness axiom.

However its elements do not have multiplicative inverses (except $1$ and $-1$) and so it is not a field.

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