Cfrgtkky

I'd like to know a sufficient condition for guaranteeing the following result:

If $f$ is uniformly continuous on $A$ and is uniformly continuous on $B$, then $f$ is uniformly continuous on $A cup B.$

In general, the above is false. But, for some cases, it is true.
For example, $f$ is uniformly continuous on $[0,1]$ and is uniformly continuous on $[1,2],$ then, clearly, $f$ is uniformly continuous on $[0,1] cup [1,2]=[0,2].$

Are there any result to extend the above example for functions from a metric space to a metric space?

Please let me know if there are any comment or answer for the question.

Thanks in advance!

OK so here was my first attempt: 'No further conditions are necessary. The definition of uniform continuity is this. Let $epsilon>0$. Then there exists $delta>0$ such that $|x-y|<delta Rightarrow |f(x)-f(y)|<epsilon$. By assumption there is a $delta$ for interval $A$, and another for $B$, so pick the lower of the two.'
Not so, because $x$ could be from $A$ and $y$ from $B$.

If $A$ and $B$ are distance at least $d$ apart, i.e. the inf of the distance between a point in $A$ and one in $B$ is $d>0$, then you are OK, because if $delta<d$ then the points $x,y$ (as in the above argument) must be both in $A$ or both in $B$.

If $A,B$ have a point in common then again you are OK because you can use the triangle inequality: let $zin A cap B$, and then $|f(x)-f(y)| < |f(x)-f(z)| + |f(z)-f(y)| < epsilon/2 + epsilon/2$, etc.

Obviously, if $A,B$ are compact, then $f$ is continuous on $Acup B$, hence uniformly continuous by Heine-Cantor, but that is a little too obvious.

If $A,Bsubseteq X$ are closed, then we know $f$ is continuous on $Acup B$. For uniform continuity, we can impose something like: there exists $delta>0$ such that every $xin Acap B$ has either $B_delta(x)subseteq A$ or $B_delta(x)subseteq B$. (This is like what we did in the elementary proof that if $fin C(mathbb{R})$ with $lim_{xtopminfty}f(x)$ exists then $f$ is uniformly continuous.)