# How do you prove a disc is open?

$$D = \{ z \in \mathbb{C} \ : \ |z| < \frac{1}{2} \}$$

It seems a disc is open or closed whether it has a < or <= but I don't know how to prove this. All examples have slightly different proofs so I've got no clarity.

-
What is your definition of open? – Qiaochu Yuan Jan 13 '11 at 15:35

The complex numbers form a metric space, meaning we have a function that say how far two points are from one another. In this case, $D$ is all the points which are not further than $\frac{1}{2}$ from $0$.

Metric spaces are very well behaved spaces, and many things are simplified in them. For example, to say whether or not a set is closed you can check if every convergent sequence inside the set has a limit inside the set. If you want to see if a set is open, you can verify that by choosing an arbitrary point and show there is a positive distance $d$ such that all points closer than $d$ are within the set, i.e. there is an open ball around your point which is completely contained in the set.

Another method is to check the complement, namely in some cases it is simpler to check if the complement is closed, rather than if the set is open; and vice versa.

In this case, however, it's easier to verify directly. That is, if $z_0$ is a complex number such that $|z_0|=r<\frac{1}{2}$ then we can take $d=\frac{\frac{1}{2}-r}{3}$ and show that if $|z-z_0|<d$ then $|z|<\frac{1}{2}$.
(which I am leaving as an exercise to the reader)

-

Think of $C$ as $R^2$ with the Euclidean metric. Then a disc is not open if it "has a <="! In that case it is closed.

Open balls (or discs if you like) in $R^2$ are open if "they have <".

That's a rule of thumb. Now you need to prove it. You do so by using the definition, that

(DEF) A set B (in this case a ball) in a metric space is open, if for each point $b \in B$ you can draw a ball $O$ around $b$ such that $b \in O \subset B$.

The proof starts by saying "Let $x$ be any point in $D$..." Then you notice that $x$ lies on the line that goes through $x$ and the centre of $D$. Consider the distances $d_1$, the distance between x and the centre and $d_2$, the distance between $x$ and the border of your set D. If you now draw a ball around $x$ of radius $r = min(d_1, d_2)/2$, this ball will be contained in $D$ and it will contain $x$ (obviously, since $x$ is its centre). Therefore you proved (DEF) and you're done.

As for proving that a ball "<=" is closed: to prove a set is closed you prove that its complement is open.

-

Hint:

Try to find some continuous function $f:\mathbb{C}\to\mathbb{R}$ such that the open disc is the inverse image $f^{-1}(U)$ for some open set $U\subset\mathbb{R}$.

-