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Let $f$ and $f_n$, $n = 1, 2, 3, \dots$, be defined on an open set $\Omega \subset \mathbb{C}$.
We'll say $f_n \to f$ uniformly on compacta if $f_n \to f$ uniformly on $K$ whenever $K$ is a compact subset of $\Omega$.

Question: Is it correct that the above condition is equivalent to the following condition (which at first glance seems a little weaker).

If $K$ is compact, and $K \subset B_r$ where $B_r \subset \Omega$ is an open disk, then $f_n \to f$ uniformly on $K$.

I think it should be. To show the second implies the first:
Cover $K$ (in the first condition) with discs which when doubled in radius are contained in $\Omega$, use compactness, look at the closure of the discs, use the second condition to get uniform convergence on the closed disks, then use the fact that $K$ is contained in the union of finitely many of them.

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Yes, that is right, as is your argument for it.

If a sequence converges uniformly on all sets belonging to a family $\mathscr{F}$ of subsets of $\Omega$, then it converges uniformly on all sets that are subsets of finite unions of members of $\mathscr{F}$. Thus, for example, compact convergence and locally uniform convergence coincide on locally compact spaces.

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