There is a problem when I am solving this question:-
Suppose $a<b<c<d$. Prove that if $f$ is uniformly continuous on $(a,b)$ and on $(c,d)$ then $f$ is uniformly continuous on $(a,b)\cup(c,d)$.
I solve the question like this:
$\forall \epsilon>0$.
As $f$ is uniformly continuous on A, then $\exists\delta_1>0\ni\forall x,y\in (a,b),|x-y|<\delta_1\implies|f(x)-f(y)|<\epsilon$
Also, $f$ is uniformly continuous on (c,d), then $\exists\delta_2>0\ni\forall x,y\in (c,d),|x-y|<\delta_2\implies|f(x)-f(y)|<\epsilon$
Take $\delta$ = $\min(\delta_1,\delta_2),\forall
x,y\in(a,b)\cup(c,d),|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$.
But when I see the solution, it is given as
Take $\delta$ = $\min(\delta_1,\delta_2, c-b)$.
Then $\forall x,y \in (a,b)\cup(c,d)$, $|x-y|<\delta \implies x,y \in (a,b) \text{ or } x,y \in (c,d),\text{ and } |x-y|<\delta_1\text{ and }\delta_2$
$\implies|f(x)-f(y)|<\epsilon$
$\therefore f$ is uniformly continuous on $(a,b)\cup (c,d)$.
But why they take $c-b$ in the expression of $\delta$?
And how does it guarantee that $x,y\in(a,b)$ or $(c,d)$ not something like $x\in(a,b)$ and $y \in (c,d)$ or vice versa?
Why we can't take $x\in(a,b)$ and $y\in(c,d)$ to prove uniform continuity on $(a,b)\cup(c,d)$?
Best Answer
Suppose $x \in (a,b)$ and $y\in (c,d).$ Then $x<b$ which gives $-x>-b$ and $y>c.$ So $|x-y|=y-x>c-b.$ Hence taking $\delta\leq c-b$ guarantees that both $x,y$ are in the same interval.
The reason you can't have both in different intervals is that to apply the two implications you have, you need both have them to be in the same interval. If $x \in (a,b)$ and $y \in (c,d),$ then there is no condition that guarantees $|f(x)-f(y)|<\epsilon.$