Show that function $f$ need not be uniformly continuous on $A ∪ B$ If $A∩B=\emptyset$ and $A$ is not compact

Question

Show that function $f$ need not be uniformly continuous on $A ∪ B$

If $A∩B=\emptyset$ and $A$ is not compact and $A$ and $B$ are closed sets.

And $f$ is uniformly continuous on both $A$ and $B$

I believe I need to use the distance between these two sets when trying to solve this question , because I proved a similar one (opposite) using the fact that if $A$ was compact, distance between these two sets would be positive.

Here $A$ is not compact, but both sets are disjoint and closed. I couldn't come up with an argument about their distances this time.

I tried thinking about a subsequence converging. Since neither of $A$ and $B$ are bounded, not all sequences $a_n$ and $b_n$ where $a_n\epsilon A$ for all $n$ and $b_n\epsilon B$ for all $n$ have a limit point. Maybe an argument from here could be derived but couldn't think of any again.

Could someone help me out? Thanks,

Answer 1

Borrowing from this answer let $A=\mathbb N$ and let $B=\{n+\frac{1}{n}| n\geq 2\}$. Clearly we can find points of $A$ and $B$ that are arbitrarily close. So if we define $f\in A\cup B$ such that $f(x)=1$ if $x\in \mathbb N$ and $f(x)=0$ otherwise then clearly $f$ is constant on $A$ and $B$ but not uniformly continuous on $A\cup B$.