Show that a bijective function so that the following holds doesn't exist

Question

Let $h: (\mathbb{R},d_2) \to (\mathbb{R},d_1)$ be a function where $d_1$ is the usual metric and $d_2: \mathbb{R} \times \mathbb{R} $ is a metric given by $d_2(x,y):=\min\{1, |x-y|\}$.

Show that there doesn't exist a bijective function $h$ so that $d_1(h(x),h(y))=d_2(x,y)$.

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I've tried to think about this question in various ways but nothings seems to get me going. I figured I would only have to show that if such a function exists it can't be injective or surjective. I also thought of showing that the domain and range aren't of equal size. But I wouldn't even know where to start with that. Could somebody help me out?

Answer 1

Suppose you had your function $h$ and let's say $h(0)=P$. Then, for all $y\in \mathbb R$ we must have $$d_1(P,h(y))=\min(1,|y|)$$ of course there are infinitely many $y$ such that the right hand is $1$, but there are only two real numbers $Q$ which satisfy $d_1(P,Q)=1$ (namely $Q=P\pm 1$).

Answer 2

For all $(x,y)$ one has $d_2(x,y)\le 1$ hence, because of $d_1(h(x),h(y))=|h(x)-h(y)|$ we have not the proposed equality when $|h(x)-h(y)|\gt 1$ (if $h$ is a bijection clearly there exist $x,y$ such that $|h(x)-h(y)|\gt 1$)

Answer 3

Assume that $h$ is bijective. Consider the sequence $x_n=h(n)$ for $n\in \mathbb{N}$. It follows that for $n>0$ $$|x_0-x_n|=d_1(h(0),h(n))=d_2(0,n)=1.$$ Get a contradiction by using the fact that $x_n$ are distinct points and $|x_0-x_n|=1$ implies that $x_0=x_n\pm 1$.