1. ## Open Sets

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

1. E is empty or
2. $\forall a \in E, \exists r > 0$ such that $D(a,r) \subseteq E$

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

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

We want to prove the statement

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

Proof:

Let $b \in D(a,r)$. Then $|b-a| < r$.
Choose $s = r - |b-a| > 0$ since $|b-a| < r$.

Let $z \in D(b,s)$. Then $|z-a| \le |z-b| + |b-a| < s + |b-a| = r - |b-a| + |b-a| = r$.
Hence $z \in D(a,r)$ and we conclude that $D(b,s) \subseteq D(a,r)$.

2. looks good to me

3. Originally Posted by 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: $E \subseteq \mathbb{C}$ is open $\iff$

1. E is empty or
2. $\forall a \in E, \exists r > 0$ such that $D(a,r) \subseteq E$

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

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

We want to prove the statement

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

Proof:

Let $b \in D(a,r)$. Then $|b-a| < r$.
Choose $s = r - |b-a| > 0$ since $|b-a| < r$.

Let $z \in D(b,s)$. Then $|z-a| \le |z-b| + |b-a| < s + |b-a| = r - |b-a| + |b-a| = r$.
Hence $z \in D(a,r)$ and we conclude that $D(b,s) \subseteq D(a,r)$.
The idea of your proof should be clear graphically. Draw a disc without its boundary, call it $D$ with center $d$ and radius $r$. Let $x\in D$ be arbitrary. Then, really all you're claiming is that the open disc with radius $r-\ell\left(\overline{dx}\right)$ is completely contained within $D$. But, this disc is merely tangent to the outer disc and since it doesn't include its boundary it follows that its contained entirely within $D$.

This was not meant to be rigorous, but a geometric reason why this is true.

P.S. Just because the adjective open is applied to open ball doesn't imply that it's open. Be careful of what you call a tautology.

4. But in a metric space, an open ball is an open set.
I don't understand why you're trying to prove a definition ?

5. Originally Posted by Moo
But in a metric space, an open ball is an open set.
I don't understand why you're trying to prove a definition ?
In my limited experience, it is often not taken that an open ball is an open set. You call an open set just any set such that for each point in the set there exists an open ball around it entirely contained in the set. Nothing in there made it neccessary that the open ball was open. But, as Slevvio showed it is not hard to prove that giving an open ball its name is justified.