A function $f(x)$ is defined $\forall x \in \mathbb{Q}$ and $ x \in (0,1)$ as $f(x) = q $ where $x = (p/q)$ in simplest form. Find $\sup_{\tau} \quad \min_{a} \quad f(a) + f(\tau-a)$ where $ a \in \mathbb{Q}$ and $ a \in (0,1)$ and $ \tau \in (\alpha,\beta)$ where $ 0< \alpha < \beta <2 $ and $ \alpha,\beta \in \mathbb{Q} $. $\sup$ and $\min$ are taken over only those $x$ where $f(x)$ is defined. Is the supremum infinity ?
EDIT
The supremum may or may not depend on $\alpha$ and $\beta$...this has not been ruled out.
EDIT 2
I will try something !
Consider $\tau = l/m$ where $l,m \in \mathbb{N}$ in simplest form and $\tau \in \mathbb{Q}\cap(0,2)$.
what is minimum possible value of $q$ where $\tau-a = p/q $ written in simplest form such that $a \in \mathbb{Q} \cap (0,1)$ and $(\tau-a) \in (0,1)$.
As Yuval Filmus suggested in his answer...
one way to choose $a$ is that it has same denominator (in simplest form) as that of $\tau$ and numerator $n$ such that $l-n$ is relatively prime to $m$, (If such a choice is possible under given conditions), then $f(a)+f(\tau-a)$ becomes $2m$.
The question is whether such a choice leading to $f(a)+f(\tau-a) = 2m$ is the minimum possible value ? If not I request you to give a counter example.
If it is, then the supremum (over $\tau$ varying in any arbitrary open interval contained in $(0,2)$) is $\infty$ as we can increase $m$ arbitrarily by varying $\tau$ in any arbitrary open interval $(\alpha,\beta)$.
EDIT 3
small change in Question...instead of $0<\alpha<\beta<2$ we have $-1<\alpha<\beta<1$
this change is not made with any intention of bringing out a certain solution.Also its influence on the solution is hoped to be very small.Inconvenience is regretted.