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Show that if $f$ and $g$ are uniformly continuous on $A \subseteq\mathbb{R}$, then $f + g$ is uniformly continuous on $A$.

How do I approach this question?

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Hint: Naturally $(f+g)(x) = f(x) + g(x)$ for all $x \in A$. With this in mind, notice that

$$|f(x) + g(x) - f(y) - g(y)| \leq |f(x)-f(y)| + |g(x) - g(y)|$$

Use the definition of $f,g$ being uniformly continuous.

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Hint: If you can come up with $\delta_g$ and $\delta_f$ that guarantee that $f$ and $g$ are within $\epsilon / 2$ of target function values when the argument of $f$ and $g$ is within $\delta_g$ and $\delta_f$ of a target argument respectively, then what happens if you put $\delta = \min(\delta_g,\delta_f)$ for the function $f + g$, and analyze how far within the target value of $f+g$ you are for that?

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