Ytukyg

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)

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

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

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 $xin[1,4]$ and $yin[4,5]$ are such $|x-y|<delta$ then

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

thus

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