Open disc in $\mathbb{C}$ with arbitrarily large radius

Question

I was wondering whether it is accurate/allowed to assume that an open disc in $\mathbb{C}$ centred at the origin with an arbitrarily large radius can be considered equal to $\mathbb{C}$ itself.

Thank you in advance!

Answer 1

I suggest you stick to the strict logical truth on this issue.

It is certainly true that $$\bigcup_{r>0} B_r = \mathbb C $$ where $B_r = \{z \in \mathbb C \,\mid\, |z|<r\}$.

It does not follow that there exists $r>0$ such that $B_r=\mathbb C$. And on top of that, it's just plain false: for any $r>0$, the number $z=(r+1) + 0i$ is in $\mathbb C$ but not in $B_r$.

Answer 2

Your question is unclear, but in a comment you amplify it to mean, is there a sense in which $\lim_{r\to\infty}D_r(0)=\mathbb{C}$?

It is obviously true that $\bigcup_r D_r(0)=\mathbb{C}$. So you could define the limit of a family of subsets to be its union. However, people generally don’t do that because it’s clearer just to explicitly state what you mean. In some contexts the “limit” you want might be the intersection, and in still other contexts something else.

If this question arose while trying to prove some theorem, better to ask about that.