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**slevvio** I worked this out myself and I was wondering if somebody could help me and see if I have made any mistakes. It's a bit of a tautology since an open disc is called an open disc but I wanted to prove this from my definition

Definition: $\displaystyle E \subseteq \mathbb{C} $ is open $\displaystyle \iff $

1. E is empty or

2. $\displaystyle \forall a \in E, \exists r > 0 $ such that $\displaystyle D(a,r) \subseteq E $

where $\displaystyle D(a,r) = \{ z \in \mathbb{C}: |z-a| < r \} $ (open disc).

So let's prove that D(a,r) itself is open!

We want to prove the statement

$\displaystyle \forall b \in D(a,r), \exists s > 0 $ such that $\displaystyle D(b,s) \subseteq D(a,r) $

Proof:

Let $\displaystyle b \in D(a,r)$. Then $\displaystyle |b-a| < r$.

Choose $\displaystyle s = r - |b-a| > 0 $ since $\displaystyle |b-a| < r $.

Let $\displaystyle z \in D(b,s)$. Then $\displaystyle |z-a| \le |z-b| + |b-a| < s + |b-a| = r - |b-a| + |b-a| = r$.

Hence $\displaystyle z \in D(a,r) $ and we conclude that $\displaystyle D(b,s) \subseteq D(a,r) $.