Confusion about $\mathbb Q$ subset suprema? This one is very confusing for me, I am probably missing some essential piece of information. 
Besides the things they told us in school, I also reassured myself with these: Is every real number the supremum of a set of rational numbers? and Is there a rational number between any two irrationals? . Though the latter probably does not add anything to my confusion. 
I found proofs that there are rational numbers between any two irrational numbers. And also every irrational number $\iota$ can be represented as a supremum of a subset of Rationals $S=\{ x \in  \mathbb Q  | x< \iota \}$. And that is the problem for me, because irrational numbers with ordering $\leq$ form a linear ordering. Therefore the map $f:\mathbb I \text{(irrationals)}\to 2^{\mathbb Q}$ given by already mentioned rule should implicate that $\iota_1<\iota_2 \iff f(\iota_1)\subset f(\iota_2)$. They cannot have same elements and there must be inclusion in way or another, because I take all the rational numbers smaller than some upper bound. This is the part that I think is probably not correct, but I do not see why at this moment.  
However that would mean there is a set of subsets of $\mathbb Q$ of uncountable size forming a linear ordering, which is impossible, because that would require an uncountable set to begin with, right? 
Where is(are) the mistake(s) in reasoning and why? And also, which is probably obvious, I do not know much about the construction of the suprema, so there might be another key information explaining this seemingly contradictory conclusions.
 A: Yes, this does seem unintuitive.
You have correctly deduced that the Dedekind cuts for the reals form an uncountable subset of $\mathcal P(\mathbb Q)$ which is totally ordered by inclusion.
This seems like a paradox at first, because each time we reach a larger subset, there must be an element of $\mathbb Q$ that was not there before but is there now. And there are not enough elements of $\mathbb Q$ that each of the uncountably many Dedekind cuts can get its own new rational that is responsible for this cut alone being larger than its predecessors.
The reason why this argument doesn't actually work is that the reals do not arise "one by one". When you're looking at one real $x$, there is not a unique "previous" real $y$ that you can compare it to and find a "new" rational that belongs to $x$ specifically.
This objection can feel somewhat unsatisfying, because it doesn't really tell you what is going on -- it just means that the counterargument you have in mind cannot be made rigorous (and is in fact not valid).
Ultimately, I think, this is just one of the cases where one has to learn to live with the fact that our intuition about finite situations leads us astray when we apply it to infinite sets. We can only point to John von Neumann's quip

Young man, in mathematics you don't understand things. You just get used to them.

and say you have to get used to the fact that when $X$ is infinite, $\mathcal P(X)$ can have totally ordered subsets that are larger than $X$ itself.
A: I believe your misinterpretation comes when you ask

However that would mean there is a set of subsets of $\mathbb{Q}$ of uncountable size forming a linear ordering, which is impossible, because that would require an uncountable set to begin with, right?

since everything before then is correct.
"A set of subsets of $\mathbb{Q}$" is the same thing as a subset $X$ of $2^{\mathbb{Q}}$, and since $2^{\mathbb{Q}}$ is uncountable (in fact, since $\mathbb{Q}$ has the same cardinality as $\mathbb{N}$, we know that $2^{\mathbb{Q}}$ has the same cardinality as $2^{\mathbb{N}}$ or $\mathbb{R}$) there is no reason why $X$ can't also be uncountable.
Your worry would be correct if you were asking about "A subset of $\mathbb{Q}$ of uncountable size forming a linear ordering," since that would be impossible. Fortunately, the set you begin with is $2^{\mathbb{Q}}$, not $\mathbb{Q}$, so you do have an uncountable set to begin with as required.
