An uncountable famliy in $2^\omega$ linearly ordered by inclusion without passing to $2^{\Bbb Q}$. I was solving the problem of finding an uncountable, linearly ordered subset of $2^\omega$ ordered by inclusion. I was having a lot of trouble before I realized that I can substitute anything countably infinite for $\omega$. $\Bbb R$ defined by Dedekind cuts is a subset of $2^{\Bbb Q}$ linearly ordered by inclusion. Bijections agree with inclusion, so I'm done.
But I don't like this solution one bit. It's just a silly trick in my opinion, and I hoped for some illumination on the structure of $2^\omega$ when I was thinking about it. Before I realized this could be done by passing to $2^{\Bbb Q}$, I tried to think about it with Cantor's diagonal argument in mind. I was trying to construct (or define, perhaps using Zorn's lemma) a set of binary sequences that would satisfy the problem's requirements. I failed, but I would like to know if it can be done.
More generally, I would like to know if there is a solution to this problem that doesn't have a bijection jumbling everything up in it. Something where you can easily see the example in $2^\omega$
 A: Probably you will consider this another silly trick. Since we can substitute any countably infinite set for $\omega$, let's use $S=\{x+iy:x,y=1,2,3,\dots\}$, the Gaussian integers in the first quadrant. For each angle $\theta\in(0,\frac{\pi}2)$ let $S_{\theta}=\{z\in S:0<\arg z<\theta\}$.
Here's another silly trick, using $\omega$ itself: for real $t>0$ let $A_t=\{2^x(2y+1):x,y\in\omega,\;y<tx\}$.
By the way, I don't consider the bijection between $\mathbb Q$ and $\omega$ a silly trick.
A: Here's the best way I can think of to "fake" the Dedekind construction in $2^\omega$ directly without jumbling stuff up too much.
Split $\Bbb N$ into chunks $A_n=[2^n,2^{n+1})\cap\Bbb N$ for $n\in\Bbb N$. So the first chunk has 1 element, the second has the next two elements, the third has four, and so on. I propose to encode a real $x\in[0,1]$ by using each chunk as an approximation of the interval, of increasing refinement. Specifically,
$$f(x)=\bigcup_{n\in\Bbb N}[2^n,2^n(1+x))\cap\Bbb N.$$
Clearly this is totally ordered w.r.t. inclusion, and it is also obviously injective and hence uncountable, because for any two different reals there is a refinement level (some $A_n$) such that the two reals are visibly separated.
    
In the figure above, the light-colored boxes are the numbers that are in the set. In this case, this is the initial segment of the encoding of $x=1/3$.
