I was wondering if there was any isometry from $\mathbb{R^2}$ to $\mathbb{R}$, using the Euclidean metric?
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no, in the real line, there are not three numbers at a common distance of $1$ apart for each pair. In the plane, there are equilateral triangles
Will Jagy
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could you explain what my teacher means then by this question , or is it a trick question .....Let X be a set and (M, d) be a metric space. Let f : X→ M be an injective map. Then we define f∗d := d ◦ (f × f): X × X →(0,inf), i.e. f*d(x, y) = d(f(x), f(y)) for x, y ∈ X. Let d denote the standard euclidean metric on R^k.Is there an injective map f : R^2 → R such that d^2 = f∗d^1? Hint: Look at a triangle in R^2 – excalibirr Oct 17 '17 at 00:03
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@jamiemma1995 your $f^\ast$ is pulling back the metric on $M$ to a metric on $X.$ Indeed, it is forcing $f$ to become an isometry. This means that there is no such map $f$ for $X = \mathbb R^2$ and $M = \mathbb R.$ No trick. – Will Jagy Oct 17 '17 at 00:13
