TITLE: Prove that there exist uncountable family of subsets of $\mathbb{N}$ which is linearly (totally) ordered by relation of inclusion.
QUESTION [5 upvotes]: Prove that there exist uncountable family of subsets of $\mathbb{N}$ which is linearly (totally) ordered by realtion of inclusion.
I do not know how to prove it. Actually I cannot concieve that such family of set can be uncountable. I would be thankful for help

REPLY [4 votes]: Hint: Fix a bijection $f \colon \mathbb N \to \mathbb Q$. For each $x \in \mathbb R$ let
$$
C_x := \{ q \in \mathbb Q \mid q \le x \},
$$
where $\le$ is the natural order of $\mathbb R$. Show that $(\{ C_x \mid x \in \mathbb R \}; \subseteq )$ is a linear order, that $C_x \neq C_y$ for $x \neq y$ and conclude, by pulling the $C_x$ back via $f$ $(\dagger)$, that there is an uncountable family $\{ D_x \mid x \in \mathbb R \}$ of subsets $D_x \subseteq \mathbb N$ which is also linearly ordered by $\subseteq$.
$(\dagger)$ I'm intentionally vague here.