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Let $f: X \rightarrow Y$ be a function on metric spaces. Let $E_{1}, ..., E_{n}$ be a finite collection of subsets of $X$, a metric space, such that $d(x,y) >1$ whenever $x \in E_{i}$ and $y \in E_{j}$ with $(i \neq j)$. Show that if $f$ is uniformly continuous on each of the sets $E_{i}$ independently, then it is uniformly continuous on $\bigcup_{i=1}^{n}E_{i}$.

Starting: Since $f$ is uniformly continuous on each $E_{i}$, we know that to every $\epsilon > 0$ there corresponds a $\delta_{i} > 0$ such that $d(f(x), f(y)) < \epsilon$ for all $x, y \in E_{i}$ such that $d(x,y) < \delta_{i}$.

The challenge is to come up with a $\delta >0$ that does what we want. Let's try: $\delta = \min \{\delta_{1}, \delta_{2}, ..., \delta_{n}\}$.

Let $x, y \in \bigcup_{i=1}^{n} E_{i}$. If $x, y \in E_{i}$, then our $\delta$ finishes the job.

Suppose $x \in E_{i}$ and $y \in E_{j}$. Then $y$ is an isolated point of $E_{i}$. Since $f$ must be continuous at any isolated point, we know that there exists a $\delta^{*} > 0$ such that for all $x \in E_{i}$ with $d(x,y) < \delta^{*} \implies d(f(x), f(y)) < \epsilon$.

I can go back and make an adjustment to my $\delta$ by considering $\delta = \min \{\delta_{1}, ...., \delta_{n}, \delta^{*} \}$. But I am not certain about whether this choice of $\delta$ depends only on the $\epsilon$ especially the $\delta^{*}$, which is really my question here. I will welcome any other ideas.

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It seems there is indeed a problem with your proof. You have to be able to give an immediate $\delta$ for a given $\epsilon$. Your $\delta$ seems to depend on other things than $\epsilon$. Use $$\delta =\min\{\delta_1, \dots \delta_n,1\}$$ instead.

If $d(x,y)< \delta$, then $x$ and $y$ must lie in the same $E_i$ (since $d(x,y) < 1$) and you will be done

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