# A set $\Omega$ is bounded if there exists $M > 0$ such that $|z| < M$ whenever $z \in \Omega$. The set $\Omega$ is contained in some large disc.

My complex analysis textbook says the following:

A set $$\Omega$$ is bounded if there exists $$M > 0$$ such that $$|z| < M$$ whenever $$z \in \Omega$$. In other words, the set $$\Omega$$ is contained in some large disc.

$$|z| < M$$ is an open disc centred at the origin, since an open disc centred at the point $$z_0$$ is defined as $$D_r(z_0) = \{ z \in \mathbb{C} : |z - z_0| < r \}$$; and so we have that a bounded set is $$\{ z \in \Omega : |z| < M \} \subseteq \Omega$$. In other words, the definition is saying that a set is bounded if every point in that set is in an open disc of radius $$M$$ centred at the origin. Am I correct in my understanding of this?

Therefore, this definition of a bounded set is specific to a disc centred at the origin.

Topologically speaking, I don't understand how this is a sensical definition for bounded sets? It seems to me that a definition of bounded sets needs to be more generalisable, rather than anchored to some arbitrary point, such as the origin.

I would greatly appreciate it if people could please take the time to clarify this.

Notice that if a set is contained in $$D_r(z_0)$$, then that set is contained in $$D_{r+|z_0-z_1|}(z_1)$$ for any $$z_1$$. In other words, the 'anchor' is irrelevant.
• @ThePointer: what this answer is saying is that if a set is contained in some disk centered at $z_0$, then we can translate $z_0$ to any other $z_1\in\mathbb C$ and adjust the radius accordingly. The center of the disk (or the "anchor" point) is irrelevant since adding two finite numbers (the radius plus the distance to the origin, for example) will still be finite. Since the center of the disk isn't important, it is just convenient to let the center be $0$ and avoid having to care about $z_0$. Jun 11 '19 at 4:25