# If $f$ is uniformly continuous on two open sets with a non-empty intersection, then $f$ is uniformly continuous on their union

There is a problem when I am solving this question:-

Suppose $$a. Prove that if $$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 $$(a,b)\cup(c,d)$$?

• Just wanted to mention that this is very well written for a first question. +1 May 18 '20 at 19:03

Suppose $$x \in (a,b)$$ and $$y\in (c,d).$$ Then $$x which gives $$-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 $$|f(x)-f(y)|<\epsilon.$$

• Great explanation!
– Iti
May 18 '20 at 10:00
• If $f(x)=1,\;\forall x\in(a,b)$ and $f(x)=0,\;\forall x\in(c,d)$. Then if we take x in (a,b) and y in (c,d) then $|f(x)-f(y)|=1$, so is not uniformly continuous. So we take into account $\delta<c-b$. Doesn't it?
– Iti
May 18 '20 at 10:05
• @Iti $f$ is still uniformly continuous on $(a,b) \cup (c,d),$ since it is uniformly continuous on each of the intervals. You are required to show that when $x,y$ are very close to each other, then so are $f(x)$ and $f(y).$ It doesn't matter what happens when they aren't close (which is the case when $x,y$ are in separate intervals). May 18 '20 at 10:13
• Ok. Now I have understood. But I have one more question, if b=c , i.e. (a,b)U(b,d) and function is defined as above in the first comment, then what should be the constraint on $\delta$ to keep x,y in same interval because if we use $\delta< c-b=0$, it will not right?
– Iti
May 18 '20 at 10:25
• @Iti If $b=c,$ then the function you defined is not uniformly continuous on $(a,b) \cup (b,d).$ Indeed for any $\delta>0,$ take $x=b-\frac{\delta}{4},y=b+\delta/4,$ then $|x-y|<\delta$ but $|f(x)-f(y)|=1.$ So your example is a great counter-example to show that it is important to take $b<c$ in the statement of the theorem. May 18 '20 at 11:02

Your proof is not correct because you can have $$x \in (a,b)$$ and $$y \in (c,d)$$ with $$|x-y| <\delta$$. This cannot happen if you make $$\delta . ( Indeed $$x \in (a,b), y \in (c,d)$$ implies $$y-x >c-b$$ because $$y>c$$ and $$x ).

• But in the solution given in book they also include (c-b) in the expression $\delta$, I haven't understood it
– Iti
May 18 '20 at 9:44
• See my edited answer. May 18 '20 at 9:47