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

Question

Let $X\subset 2^{\mathbb Z}$ be the set that consists of subsets of $\mathbb Z$ that are bounded above. That is to say, $A\in X$ iff $A\subset\mathbb Z$ and $\exists\max A$. Show that the set $X$ has the property of Supremum; namely, that there is a least upper bound for any bounded set, of bounded above subsets of $ \mathbb Z$. The idea is to provide an explicit construction of the supremum, in at most infinite denumerable steps for any case. Of course, the first step is to define a proper order which is already worked out. The order is as follows: Let $A,B\in X$ two different sets, which makes the symmetric difference non empty. We say $A<B$ iff $\max A\triangle B\in B$. Otherwise, if $\max A\triangle B\in A$ we say $A>B$. It is not too difficult to prove this is a linear order and that every two elements are comparable. So with this, we try to build the supremum of $X$. Why? Because this set is isomorphic to $\mathbb R$ in terms of order, and an explicit isomorphism can be given using the binary expansion of real numbers.

Answer 1

Let's restrict attention to nonempty sets of integers, for simplicity. (Also, because $\emptyset$ is minimal according to your ordering, which I don't think you want.)

For clarity I'll write "$\triangleleft$" for your ordering to distinguish it from the usual ordering $<$ on $\mathbb{Z}$.

If $A\subseteq X$ is nonempty and $\triangleleft$-bounded above, then it does indeed have a supremum, as follows. The key observation is that a $\triangleleft$-bounded set has a maximum integer occurring in any of its elements.

• Let $z_0=\max(\bigcup A)$ and let $A_0=\{B\in A: z_0\in B\}$.

• Having defined $z_i$ and $A_i$, let $z_{i+1}=\max((\bigcup A_i)\setminus\{z_i\})$ if it exists, and $z_{i+1}=z_i$ otherwise, and let $A_{i+1}=\{B\in A_i: z_{i+1}\in B\}.$

It's not hard to show that $\sup(A)=\{z_i: i\in\mathbb{N}\}$.

That said, it's not yet obvious to me that in fact you get something order-isomorphic to $\mathbb{R}$. The supremum property alone doesn't characterize $\mathbb{R}$ as an ordering (think about an arbitrary ordinal). EDIT: In fact, unless I'm missing something they're not order-isomorphic: there is nothing strictly $\triangleleft$-between $\{..., -3,-2,-1\}$ and $\{0\}$.