Metrics that generate the same convergent sequence.

Question

$(*)\ c_1 d_1(x,y) \le d_2(x,y) \le c_2 d_1(x,y)$ for some constants $c_1,c_2 > 0$.

Let $d_1 = |x-y|$ and $d_2 = \sqrt{|x-y|}$ be metrics on $\mathbb{R}$. Show that these metrics do not satisfy $(*)$ for any $c_1, c_2$.

Proof: $c_1 |x-y| \le \sqrt{|x-y|} \le c_2 |x-y| = c^2_1 |x-y|^2 \le |x-y| \le c^2_2 |x-y|^2 = c^2_1\le \frac{1}{|x-y|} \le c_2^2 = c_1\le \frac{1}{\sqrt{|x-y|}} \le c_2 = c_1\le \frac{1}{d_2} \le c_2$.

I do not know where to continue from here though, from the looks of it, I'm just proving that it does satisfy $(*)$. Does it have something to do with bounds?

Answer 1

Assume that $c_1$ and $c_2$ exist and get a contradiction by choosing $x$ and $y$ appropriately. For example, $\sqrt {|x-y|} \leq c_2|x-y|$ cannot hold for all $x,y$ as seen by taking $x=\frac 1 n, y=0$ where $n >c_2^{}$. Sim ialrly, $c_1|x-y| \leq \sqrt {|x-y|}$ is violated when $x=n,y=0$ with $n >\frac 1 {c_1^{2}}$.