For any non-empty subset $A$ of real numbers define $f_A : \mathbb{R} \to \mathbb{R}$ by, $$f_A(x)= \inf \{ |x-y| : y \in A \}$$
Let, $A,B$ be disjoint non-empty sets of real numbers such that $f_A(x)+ f_B(x) \gt 0$ for all $x \in \mathbb{R}$. Define a function $g : \mathbb{R} \to \mathbb{R}$, $$g(x) = \frac{f_A(x)}{f_A(x)+ f_B(x)}$$
Then,
$(i)$ why $g$ is continuous?
$(ii)$ give an example of two disjoint non-empty subsets of real numbers $A,B$, such that $f_A(x)+ f_B(x) \gt 0$ for all $x \in \mathbb{R}$ and for which $g$ is not uniformly continuous.
I've proved that the function $f_A(x), f_B(x)$ are continuous. So their sum is also continuous and hence $g$ being the quotient of two continuous functions whose denominator is always positive, is continuous.
For the second part I've thought of closed disjoint sets $A,B$. For example I thought of $A = \{0\}$ and $B = \{-1\}$. Then the function $g$ becomes $$g(x) = \frac {|x|}{|x|+|x+1|}$$
But I can't check if it's not uniformly continuous. Plotting it I found that there's no steep ascent or descent so it must be uniformly continuous.
Can anyone please help me with the part two??
Best Answer
Let $A=\{1, 2, 3, 4, ...\}$. Let $B=\{1+\frac 12, 2+\frac 13, 3+\frac 14, 4+\frac 15, ...\}$.
Thus for any $\epsilon$ we can find $a_i\in A$, $b_i \in B$ so that $|a_i-b_i|<\epsilon$. Thus $g(a_i)=0$, $g(b_i)=1$.