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If $f$ is uniformly continuous on $[1,4]$ and uniformly continuous on $[2,5]$ is $f$ uniformly continuous on $[1,5]$?

I know that the answer is that it is true, but I am interested as to why my proof is incorrect. What I did was:

Since

$\exists \delta_1 s.t. |x-y|<\delta_1 \to |f(x)-f(y)|<\frac{\epsilon}{2}$ for $x,y \in [1,2]$ and $\exists \delta_2 s.t. |x-y|<\delta_2 \to |f(x)-f(y)|<\frac{\epsilon}{2}$ for $x,y \in [2,5]$

Then set $\delta_3 = \min{\delta_1,\delta_2}$ and you get (by triangle inequality)

$|f(x)-f(y)|=|f(x)-f(3)+f(3)-f(y)|\leq |f(x)-f(3)| + |f(y)-f(3)|\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

as long as $|x-y|<\delta_3$ What's wrong with this proof and how can I fix it?

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  • $\begingroup$ What does $3$ have to do with it? In general $|f(x) - f(3)|$ and $|f(y) - f(3)|$ are not small. $\endgroup$
    – Robert Israel
    Nov 20, 2018 at 20:27
  • $\begingroup$ @RobertIsrael because 3 is in both intervals, so we know that there exists a delta that makes $|f(x)-f(3)|<\epsilon$, right? $\endgroup$
    – Riley H
    Nov 20, 2018 at 20:31
  • $\begingroup$ Would there be any other changes if you used continuity at 2? @hamam_Abdallah $\endgroup$
    – Riley H
    Nov 20, 2018 at 20:32
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    $\begingroup$ Here is a bizarre proof: If $f$ is continuous on $[1,4]$ and on $[2,5]$, then it is continuous on $[1,5]$ by the pasting lemma. Then, being continuous on a compact set, it is uniformly continuous. $\endgroup$
    – Sangchul Lee
    Nov 20, 2018 at 20:35
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    $\begingroup$ For the standard proof, let $\delta_1, \delta_2$ be as in your setting and let $\delta = \min\{\delta_1, \delta_2, 2\}$. If $x,y\in[1,5]$ satisfy $|x-y|<\delta$, argue that either $x,y\in[1,4]$ or $x,y\in[2,5]$ holds, so that the defining property of $\delta_i$'s kicks in. $\endgroup$
    – Sangchul Lee
    Nov 20, 2018 at 20:39

1 Answer 1

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hint

The uniform continuity at $[1,4]$ gives $\delta_1$

the UC at $[4,5]$ will give $\delta_2$.

the continuity at $x=4$ gives $\delta_3$ such that

$$|x-4|<\delta_3 \implies |f(x)-f(4)|<\frac{\epsilon}{2}$$

Take $\delta=\min(\delta_i,i=1,2,3).$

If $x\in[1,4]$ and $y\in[4,5]$ are such $|x-y|<\delta$ then

$|x-4|<\delta_3$ and $|y-4|<\delta_3$

thus

$$|f(x)-f(y)|=$$ $$=|f(x)-f(4)+f(4)-f(y)|<\epsilon$$

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