Enumeration Hierarchies 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 $\aleph_0$?

 A: 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.
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, 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 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, 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 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) Suppose the 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 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.
