Let $f: \mathbb{Q} \rightarrow \mathbb{R}$ such that: $$x \mapsto \left\{\begin{matrix} 1 & \text{if } x > \pi\\ 0 & \text{if } x < \pi \end{matrix}\right.$$ Show that $f$ is continuous in the usual metrics. Then show that if we use the metric: $\delta: \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{R} $ such that: $$(x,y )\mapsto |\frac{1}{\pi -x} - \frac{1}{\pi -y}| + |x-y| $$ then $f$ is Lipschitz continuous with the Lipschitz contant equal to $\frac{1}{4}$.
So in order to show that $f$ is continuous, here was my approach:
Let $x \in \mathbb{Q}$, suppose that $x < \pi$. Let $\epsilon > 0$, then let's put $\eta = \frac{|x - \pi|}{2}$, then $\forall y \in \mathbb{Q}$ such that $|x-y| < \eta$, we necessarily have $y < \pi$, thus $f(x)=f(y)$ and so $|f(x)-f(y)|=0<\epsilon$. In similar fashion I would reason for $x > \pi$.
Now in order to show that $f$ is Lipschitz continuous with Lipschitz constant equal to $\frac{1}{4}$, I would need to show that $|f(x)-f(y)|\leq \frac{1}{4} \delta(x,y)$. Now this is obvious if $x,y > \pi$ or $x,y < \pi$ because then $f(x)=f(y)$, thus $|f(x)-f(y)|=0$. The issue I get is when I try to show that for $x<\pi<y$ we have $|f(x)-f(y)|\leq \frac{1}{4}\delta(x,y)$.
We have $f(x)=0$ and $f(y) = 1$, thus $|f(x)-f(y)|=1$, but I don't see how $1 \leq \frac{1}{4}(|\frac{1}{\pi -x} - \frac{1}{\pi -y}| + |x-y|) $.
For $x < π < y$,\begin{align*} δ(x, y) &= \left|\frac{1}{π - x} + \frac{1}{y - π}\right| + |y - x| = \frac{1}{π - x} + \frac{1}{y - π} + y - x\\ &= (y - π) + \frac{1}{y - π} + (π - x) + \frac{1}{π - x} \geqslant 2 + 2 = 4 = 4|f(x) - f(y)|, \end{align*} so$$ |f(x) - f(y)| \leqslant \frac{1}{4} δ(x, y). $$