Prove that uniform convergence of a series in a region $G$ implies uniform convergence in any portion of $G$

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"Prove that uniform convergence of a series in a region $G$ implies uniform convergence in any portion of $G$." (Exercise 18.c PSet 15.5, Advanced Engineering Mathematics, 10th Ed.)

It's seems so obvious that I don't know what to write, but maybe I'm missing some subleties.

The definition of uniform convergence is the following

\begin{equation} \forall \epsilon>0 \exists N \forall n>N \forall z \in G |s(z)-s_n(z)|<\epsilon \tag{1} \end{equation}

Being $G_1 \subset G$ a subregion, it's clear that it's also true that

\begin{equation} \forall \epsilon>0 \exists N \forall n>N \forall z \in G_1 |s(z)-s_n(z)|<\epsilon \tag{2} \end{equation}

What should I write? The answer can't be "it's clear/obvious that".

By the way, "A region is a set consisting of a domain plus, perhaps, some or all of its boundary points." and "An open and connected set is called a domain" (kreyszig)

Thanks, Luca

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$$G_1\subset G \implies $$

$$\sup_{z\in G_1}|s (z)-s_n (z)|\le \sup_{z\in G}|s (z)-s_n(z)|$$

for each $n $.

If the right side goes to zero, then the left one goes also to zero by squeeze theorem.