The infinite union of open disks "Consider the following open disks:
$D(0,1) = \{z ∈ \mathbb{C} : \lvert z\rvert < 1\}$
$D(0,p) = \{z ∈ \mathbb{C} : \lvert z\rvert < p\}$ with $0<p<1$
i) Prove that $D(0,1) = \cup_{0<p<1}D(0,p)$
ii) Is it possible to find $0<p_1<...<p_i<...<p_n<1$, such that $D(0,1) = \cup_{i=1}^{n}D(0,p_i)$?
Justify your answer."
Right now I'm struggling to figure out where to start a rigorous proof. It just seems obvious, but I'm not sure how to put it into words
 A: For (i), we have
$x \in D(0, 1) \Longleftrightarrow \vert x \vert < 1 \Longleftrightarrow \exists p, \vert x \vert < p < 1$
$\Longleftrightarrow \exists p, x \in D(0, p) \Longleftrightarrow x \in \cup_{0 < p < 1} D(0, p); \tag 1$
as for (ii), it is false since if $p_n < \vert x \vert < 1$, $x \notin D(0, p_n)$, hence $x \notin D(0, p_i)$ for $1 \le i < n$, since $D(0, p_i) \subset D(0, p_n)$, and thus $x \notin \bigcup_{i=1}^n D(0, p_i)$.
If, on the other hand, $p_i \in (0, 1)$ is a sequence with $p_i \to 1$ as $i \to \infty$, then it is easy to see that
$D(0, 1) = \bigcup_{i=1}^\infty D(0, p_i); \tag 2$
the proof is of course very similar to those of items (i) and (ii) above.
A: i).For all $z\in \Bbb C$ we have $$z\in D(0,1)\implies |z|<1\implies |z|<(1+|z|)/2<1\implies$$ $$\implies [\,|z|\in D(0,(1+|z|)/2)\,\land \, 0<(1+|z|)/2<1 \,]\implies$$ $$\implies  \exists p\;(\;0<p<1\land |z|\in D(0,p)\;)$$ $$\implies z\in \cup_{0<p<1}D(0,p)\implies$$ $$\implies \exists p<1\;(\,z\in D(0,p)\,)\implies$$ $$\implies \exists p<1\,(|z|<p)$$ $$ \implies |z|<1\implies z\in D(0,1).$$
Therefore for all $z\in \Bbb C$ we have
$z\in D(0,1)\implies z\in \cup_{0<p<1}D(0,p)\implies z\in D(0,1).$
ii). If $0<p_1<...<p_n<1$ then $|(1+p_n)/2|=(1+p_n)/2<1$ so $$(1+p_n)/2\in D(0,1) \setminus \cup_{j=1}^n D(0,p_j)$$ because $|(1+p_n)/2|=(1+p_n)/2>p_n\geq p_j$ for each $j$ from $1$ to $n.$
