If $d(x,y)=|f(x)-f(y)|$ is a metric in $\mathbb{R}$ where $f$ is continuous also a strictly increasing function and $d_{E}(x, y) =|x-y|$ is the Euclidean metric, I want to prove that:
For every $\varepsilon>0$ and $x \in \mathbb{R}$ there is a $\delta > 0$ such $B_{d}(x,\delta) \subseteq B_{d_{E}}(x, \varepsilon)$.
Basically, someone help me giving the $\delta>0$ I was looking for in a previous question as $$\delta=\max \lbrace f(x)-f(x- \varepsilon), f(x + \varepsilon)-f(x) \rbrace.$$
So if I take $y \in B_{d}(x,\delta) $ then $d(x,y)< \delta$, then I should consider both cases for $\max \lbrace f(x)-f(x- \varepsilon), f(x + \varepsilon)-f(x) \rbrace$. If $\delta= f(x + \varepsilon)-f(x)$ I have
$$|f(x)-f(y)|< f(x + \varepsilon)-f(x).$$
But I cannot get to the point where $|x-y| < \varepsilon$ in order to end this proof. Also stuck in the case where $f(x)-f(x- \varepsilon)$. I've been trying this problem for many days and any help to finally end this proof will be really apreciated. Thanks.
The $\max$ really ought to be a $\min$.
If $y \ge x$, then $|f(x) - f(y)| = f(y) - f(x)$, so $$|f(x) - f(y)| < \delta \implies f(y) - f(x) < \delta \le f(x + \varepsilon) - f(x) \implies f(y) < f(x + \varepsilon).$$ Now, if $y \ge x + \varepsilon$, we would expect to see $f(y) \ge f(x + \varepsilon)$, since $f$ is increasing. This is not the case, so $y < x + \varepsilon$. Given $y \ge x$ was assumed, this implies $|x - y| < \varepsilon$.
Similarly, if $y < x$, then $|f(x) - f(y)| = f(x) - f(y)$, so $$|f(x) - f(y)| < \delta \implies f(x) - f(y) < \delta \le f(x) - f(x - \varepsilon) \implies f(x - \varepsilon) < f(y).$$ Assuming $y \le x - \varepsilon$ similarly contradicts $f$ being increasing, so $x - \varepsilon < y < x \implies |x - y| < \varepsilon$.