Explanation of a statement in proposition 20 in Royden " Real Analysis."

Question

The proposition and its proof are given below:

My questions are:

1-Why $\emptyset(1) = 1$?

2-why in the second picture "if $k$ is sufficiently large $x_{0}$ lies between 2 consecutive intervals in $O_{k}$"?

Answer 1

1. There are elements of $\mathcal{O}=[0,1]\setminus \mathcal{C}$ that are arbitrarily close to $1$, so $\varphi(1)=1$. For instance, $$1- \frac{1}{2}\left( \frac{1}{3^n}+\frac{2}{3^n}\right) = 1- \frac{1}{2}\cdot \frac{1}{3^{n-1}} \in \mathcal{O} \text{ for all } n \geq 1$$ since the interval $\left(1-\frac{2}{3^n}, 1-\frac{1}{3^n}\right)$ is removed from $\mathcal{C}$ for every $n \geq 1$.

2. $x_0$ is some positive distance away from both $0$ and $1$, say $\Delta$. Note that two of the intervals removed at the $k$th step are $$\left(\frac{1}{3^k}, \frac{2}{3^k}\right) \text{ and } \left(1-\frac{2}{3^k}, 1-\frac{1}{3^k}\right).$$ For sufficiently large $k$, we have $\frac{2}{3^k}<\Delta$, and so these intervals will lie on opposite sides of $x_0$. So $x_0$ is between some intervals removed at the $k$th stage, and since only finitely many intervals are removed we can selected the rightmost one that is left of $x_0$, and the leftmost one that is right of $x_0$, which will be consecutive.