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Let $a<b<c<d$ and $f: (a,d) \rightarrow \mathbb{R}$. Assume $f$ is uniformly continuous on $(a,c)$ and $(b,d)$. Prove that $f$ is uniformly continuous on the interval $(a,d)$.

My proof [EDITED]: Let $f$ be uniformly continuous on $(a,c)$ and $(b,d)$. So $\forall \epsilon_1 \exists \delta_1$ s.t. if $x_1,y_1 \in (a,c)$ and $|x_1 - y_1| < \delta_1$ then $|f(x_1)-f(y_1)| < \epsilon_1$ and $\forall \epsilon_2 \exists \delta_2$ s.t. if $x_2,y_2 \in (b,d)$ and $|x_2 - y_2| < \delta_2$ then $|f(x_1)-f(y_1)| < \epsilon_2$. From here, I'm not sure where to go.

Is my proof clear? Or is it missing some needed information? Thanks in advance for any suggestions or help.

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Nonempty bounded open intervals are not closed. Use the definition of uniformly continuous. –  Jonas Meyer Oct 25 '12 at 6:26
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and $(a,c)$ and $(b,d)$ are both compact sets... Ouch! That hurts... –  Did Oct 25 '12 at 6:31
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Just some minor suggestions for getting started: You will want to choose $\delta_1$ and $\delta_2$ based on the same $\varepsilon$, then use them to find one $\delta$ that works for the union. In doing so, you might want to keep in mind that making $\delta$ smaller never hurts. You may also want to sketch the intervals and consider cases to organize your thoughts. –  Jonas Meyer Oct 25 '12 at 6:50
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2 Answers 2

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If you have some $x_1 \in (a,c)$ and $x_2 \in (b,d)$, and you have $|f(x_2)-f(x_1)|$, since we know there is some overlap in $(a,c)$ and $(b,d)$, we can find another number $x_3 $in this overlap. We can then use this fact: $|f(x_2)-f(x_1)|=|f(x_2)-f(x_3)+f(x_3)-f(x_1)|$ and the triangle inequality. That's my intuition at least.

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That's one good approach. Another is to choose $\delta$ small enough to ensure (among other things) that $|x-y|<\delta$ implies that $x$ and $y$ are both in $(a,c)$ or both in $(b,d)$. –  Jonas Meyer Oct 25 '12 at 7:03
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You are right that the compact sets in $\mathbb{R}$ are the sets which are closed and bounded. And there is also a result on metric spaces that a continuous function on a compact space is uniformly continuous. But the other way is not true. That the function is uniformly continuous on a set does not imply that set is compact. $(a,c)$ is an open interval, and not closed or compact.

As already pointed out in the comments, you should use the definition of uniformly continuous.

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