How to prove a totally ordered set is not Dedekind complete? Take $\mathbb{Z}^{\mathbb{N}}$ to be the set of all sequences of integers. Where every element $x$ is of the following form: $x=(x_{1},x_{2},...)$ with $\forall n\in \mathbb N^*, x_{n}\in\mathbb{Z}$.
We order the set as follows: $x\lneq y\Leftrightarrow$ the first non-zero entry of $y-z$ is positive. Where we denote subtractions as term by term such that: $y-z$ corresponds to the sequence $(y_{1}-x_{1},y_{2}-x_{2},...,)$
I'm trying to prove that this totally ordered set is indeed not Dedekind complete (every non-empty subset that is bounded above has a supremum). I've checked previously that this ordering does form a totally ordered set.
I'm a bit stuck on where to begin this proof. Any advice is appreciated.
 A: It's easier to see what's going on if we rearrange things a bit. I am going to start $\mathbb{N}$ at $0$ and also rename your sequences to $a_0, a_1, a_2, \dots $, and furthermore I am going to think about a sequence as a formal power series
$$A(x) = \sum_{i=0}^{\infty} a_i x^i.$$
The reason I want to do this is that we get a conceptual interpretation of the lexicographic order in terms of formal power series. Two formal power series $A(x), B(x)$ satisfy $A < B$ iff $B - A$ is positive in the sense that its first nonzero term is positive. What this tells us is that $x$ behaves like a positive infinitesimal; it is smaller than all of the positive "ordinary numbers" $a_0$ but larger than $0$, and also larger than $x^2$, which is larger than $x^3$, which is larger than $x^4$, and so on. So the lexicographic order can be thought of as describing "formal power series in a positive infinitesimal."
So, how do we use this interpretation to produce a non-empty set bounded from above without a supremum? Consider the set
$$x, 2x, 3x, \dots $$
which consists of infinitesimals so is bounded from above by $1$. In fact it's not hard to see that the set of upper bounds is exactly the set of power series $g(x)$ such that $g(0) \ge 1$. What this means is that if $g(x)$ is any upper bound then so is the infinitesimally smaller upper bound $g(x) - x$, so there is no least upper bound. (Obviously you could run this entire argument in terms of sequences but personally I think it's less intuitive, whereas thinking about $x$ as infinitesimal immediately suggests this argument, to me anyway.)
Note that we only need a single infinitesimal to run this argument, not the extravagantly larger infinite stack of infinitesimals we actually have available. The general lesson here is that Dedekind-completeness is not compatible with the existence of infinitesimals (say, when you are trying to figure out what properties you should require $\mathbb{R}$ to have).
