I encountered the following problem: If you sort sets by the following: $$A\lt B:\Leftrightarrow \exists f:A\rightarrow B\; injective,\quad \nexists g:B\rightarrow A \; injective$$ You receive the standard order of sets. Now obviously, $$\emptyset\le\{1\}\le\{1,2\}\le...\le\mathbb{N}\le\mathbb{R}\le\mathbb{R}^\mathbb{R}\le\mathbb{R}^{\mathbb{R}^\mathbb{R}}$$ Does the continuum hypothesis imply that there are no sets "inbetween" $\mathbb{R}$ and $\mathbb{R}^\mathbb{R}$, in other words, $$\nexists A: \mathbb{R}\lt A\lt\mathbb{R}^\mathbb{R}$$ and so on ($\nexists A: \mathbb{R}^\mathbb{R}\lt A\lt\mathbb{R}^{\mathbb{R}^\mathbb{R}}$)?
Does this order continue on and on, where a $\mathbb{N}$ gets replaced by a $\mathbb{R}$ and every second step, a $\mathbb{N}$ is added?
I am asking because I always saw the (aleph-) notation $\aleph_1:=2^{\aleph_0}$ but that doesnt tell me too much about this question; Basically my question is are $\aleph_n \;n\in\mathbb{N}_0$ all of the cardinalities there are? Is $|\mathbb{R}^{...^\mathbb{R}}|=\aleph_{k}$?