Converges uniformly on an arbitrary closed disc implies on every compact subsets

Question

Suppose that we have a given sequence of functions $(f_n)_{n\geq 0}$. The goal is to show that it converges uniformly on every compact subsets of $\mathbb{C}$.

Let $R>0$ be arbitrary, and define, say, $C_R=\{z\in \mathbb{C}\mid |z-1|\leq R\}$. If one has shown that $(f_n)_{n\geq 0}$ converges uniformly on $C_R$, how do you conclude mathematically that it then converges uniformly on every compact subsets of $\mathbb{C}$?

I can't find a correct way to conclude: obviously, I would just say that $R$ could be taken arbitrarily large, but in this case, what's been shown, is that it actually converges on every compact subsets containing $1$. See the definition of $C_R$, which is a closed disc around $1$ with centre $R$.

Answer 1

Say that $K$ and $K'$ are two compact sets and $K \subset K'$. Then if $(f_n)$ converges uniformly on $K'$, it also converges uniformly on $K$.

Therefore to check that it converges uniformly on all compact subsets of the plane, you just need to give a family of compact subsets of the plane such that every compact subset of the plane is a subset of a member of the family. These disks are one such family.

Answer 2

If $K$ is any compact set then $K \subseteq \{z: |z-1| \leq R\}$ for $R$ sufficiently large. Just take $R=1+\sup \{|z|:z \in K\}$.