Is there an isometry from $\Bbb R$ to $\{(x,y) \mid x,y\in \Bbb R, y= x^2\}$

Question

I think there is not, but I do not know how. I try to start with the simplest point $0$, and assume that $f(0) = a$ then $|f(x) - f(0)|=|x|$, but I do not know how to get a contradiction.

in $\{(x,y) \mid x,y\in \Bbb R, y= x^2\}$ the metric is the standard Euclidean metric from $\Bbb R^2$

Answer 1

Hint. On the line, $a < b < c$ implies something about the distance from $a$ to $c$.

Hint. What can you say about the "triangle" inequality for three sides of the triangle with vertices $a$, $b$ and $c$?. What about the triangle with vertices $f(a)$, $f(b)$ and $f(c)$.