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?