A set $M$ of real values is said to be enumerable if there is a bijection between the elements of $M$ the elements of $\mathbb{N}$.
That definition does however not impose any restrictions on the order in which the elements of $M$ must be enumerated.


  


> 
> **Questions:**
> 
> 
> defining two enumerable sets $A$ and $B$ of real values as being isomorphic iff a strictly monotone bijection between the elements of $A$ and $B$ exists, i.e. if a pair of enumerations, one of the elements of $A$ and one of the elements of $B$, $A=\lbrace a\_i\rbrace\_{i=1}^\infty$, $\ B=\lbrace{b\_i}\rbrace\_{i=1}^\infty$, exists for which $\left(a\_{i+1}-a\_i\right)\left(b\_{i+1}-b\_i\right)\ \gt\ 0\quad \forall i\ge 0$ holds,
> 
> 
> * what is the cardinality of a maximal set of pairwise non-isomorphic enumarable subsets of $\mathbb{R}^+$
> * Is the set $\mathbb{Q}$ of rational numbers isomorphic to the set of algebraic numbers?
> * can the above notion of isomorphism between sets be generalized beyond $\aleph\_0$?
> 
> 
> 



Enumeration Hierarchies
For the first two thirds of your question, you're really just talking about **countable linear orders**. Every set $A\subseteq\mathbb{R}$ can be viewed as a linear order in a natural way (via the usual ordering of $\mathbb{R}$), and two sets are isomorphic in your sense iff they are isomorphic when thought of as linear orders. There are indeed continuum many isomorphism types of countable linear orders: one way to see this is to consider, given a function $f:\mathbb{N}\rightarrow \mathbb{N}$, the linear order $$\mathcal{L}\_f:=\mathbb{Z}+f(0)+\mathbb{Z}+f(1)+...$$ where we conflate $n\in\mathbb{N}$ with the $n$-element linear order. It's easy to show that $\mathcal{L}\_f\cong\mathcal{L}\_g\iff f=g$.


*EDIT: It's worth noting that when we replace "isomorphism" with "bi-embeddability," the answer changes drastically: [the number of bi-embeddability classes of countable linear orderings is $\aleph\_1$, provably in ZFC](https://math.stackexchange.com/a/2948318/28111)*.


Now, you might object that you're not asking about *arbitrary* countable linear orders, but rather countable linear orders *which can be embedded into $\mathbb{R}$*. But these are the same thing. Indeed, every countable linear order can be embedded into $\mathbb{Q}$. Moreover, any two countable dense linear orders without endpoints are isomorphic (so this answers your question about $\mathbb{Q}$ versus the algebraic numbers). The latter fact was [proved by Cantor](https://en.wikipedia.org/wiki/Dense_order#Uniqueness), using what is now called a "back-and-forth argument"; I don't know who first observed the former, but it's proved by basically the same idea *(or rather, half of the same idea: we only care about one direction of the map we want to build)*, that $\mathbb{Q}$ is "sufficiently saturated" to let us build embeddings inductively without ever running into problems. It's worth noting that Cantor's theorem is entirely constructive: given two countable dense linear orderings without endpoints (with enumerations), it produces an explicit isomorphism. So in particular we can actually write down an isomorphism between $\mathbb{Q}$ and the algebraic numbers, once we fix enumerations of each. However, I know of no *nice* isomorphism between the two.




---


Now, what about *uncountable* sets of reals? *([This paper by Moore and Todorcevic](http://pi.math.cornell.edu/~justin/Ftp/Baumgartner_aleph2.pdf) is relevant here.)*


Well, first of all, it's no longer true that every uncountable linear order embeds into $\mathbb{R}$. Indeed, this isn't even true for linear orders of size $\aleph\_1$: the first uncountable ordinal $\omega\_1$ does not embed as a linear order into $\mathbb{R}$ (since the rationals are dense and countable). So here we really do need to say that we're looking at linear orders of size $\kappa$ which are embeddable into $\mathbb{R}$.


It turns out that in fact the situation is **very** different from the countable situation:



> 
> **(Baumgartner)** It is consistent with ZFC that any two $\aleph\_1$-dense linear orders without endpoints which are embeddable into $\mathbb{R}$ are isomorphic.
> 
> 
> 


Here, a linear order is $\aleph\_1$-dense if between any two points there are $\aleph\_1$-many additional points. The opposite situation is also possible: 



> 
> **(Dushnik-Miller)** There are $2^{2^{\aleph\_0}}$-many nonisomorphic linear orders of cardinality $2^{\aleph\_0}$ which can be embedded into $\mathbb{R}$.
> 
> 
> 


(In fact, they actually proved something stronger.) Thus if CH holds, we get a situation completely opposite Baumgartner's.


So once we look beyond countable linear orders, the situation becomes highly dependent on set theory.


A natural question is to ask about the $\aleph\_2$-version. Itay Neeman announced a solution a few years ago - that the $\aleph\_2$-version of Baumgartner's theorem is consistent with ZFC relative to a weakly compact cardinal - but I believe that paper hasn't appeared yet. Notes on the argument can be found [here](https://settheory.mathtalks.org/wp-content/uploads/2016/11/GiltonNeeman_SideConditionsAndIterationTheorems.pdf), but I don't know the extent to which they've been vetted (and they're well beyond my comprehension); a good starting place for approaching this, I believe, is [Neeman's Notre Dame paper](https://projecteuclid.org/euclid.ndjfl/1398345784) on the "modified side conditions" approach, but it assumes existing comfort with forcing and classical side conditions.




---


It's worth mentioning one final type of structure theorem:


A basic result about countable linear orders is that every countable (infinite) linear order either has a copy of $\omega$ or a copy of $\omega^\*$ (or both) as a suborder. We say that $\{\omega,\omega^\*\}$ form a *basis* for the countably infinite linear orders.


We can ask about the existence of bases "higher up." A basis for the size-$\kappa$ linear orders is a set of linear orders, each of size $\kappa$, such that any linear order of size $\kappa$ contains one of the orders in the set as a suborder. We may reasonably ask at this point: *is there a "simple" basis for the size-$\aleph\_1$ linear orders?*


It turns out that consistently (relative to large cardinals) the answer is **yes**:



> 
> [**(Moore)**](http://annals.math.princeton.edu/wp-content/uploads/annals-v163-n2-p08.pdf) Suppose the [Proper Forcing Axiom](https://en.wikipedia.org/wiki/Proper_forcing_axiom) holds. Then there is a five-element basis for the size-$\aleph\_1$ linear orders - namely, any set of the form $$\{\omega\_1,\omega\_1^\*, C, C^\*, X\}$$ where $X$ is a set of reals of size $\aleph\_1$ and $C$ is a [Countryman line](https://en.wikipedia.org/wiki/Countryman_line) forms a basis for the size-$\aleph\_1$ linear orders.
> 
> 
> 


Note that since any uncountable linear order contains a suborder of size $\aleph\_1$, we may equally say that the set of *uncountable linear orders* has a five-element basis (and this is how Moore phrases his result in the linked paper).


As far as I know, we have no significant understanding of bases for the size-$\aleph\_2$ linear orders.


