I'm studing measure in the Bartle's book and his coments that the closed disk:
$$D=\{(x,y)\in\mathbb{R}^2 | x^2+y^2\leq 1 \}$$
Is the countable union of disjoint cell's (intervals in $\mathbb{R}^n$).
But i'm dont understand why this not contradicts the connected of the disk , because if $$D=\bigcup_{n\in\mathbb{N}}I_n$$
Then i can write $$D= I_k\cup \bigcup_{n\neq k} I_n$$, and this sets they are disjoints , not empty and your union is D , so D have an not trivial cut.