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Let $(X,d)$ be a metric space.

Let $x_0,x_1\in X$ and $p\gt0$ a constant.

I want to show that the set $\{\dfrac{1+d(x_0,x)^p}{1+d(x_1,x)^p}: x\in X\}\subset \Bbb R$ is bounded.

I was trying to use the triangle inequality but with no success: $$\dfrac{1+d(x_0,x)^p}{1+d(x_1,x)^p}\le \dfrac{1+d(x_0,0)^p + d(0,x)^p}{1+d(x_1,x)^p-d(0,x_1)^p} \le \dfrac{A +d(0,x)^p}{B-d(0,x)^p}$$

Where $A,B$ are constants.

Any idea on how to continue?

Thanks a lot!

infinity
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1 Answers1

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You are on the right track: As a start, for $p = 1$, compute

$$\frac{1 + d(x_0,x)}{1+d(x_1,x)} \leq \frac{1+d(x_0, x_1) + d(x_1,x)}{1+d(x_1,x)} = \frac{C + d(x_1,x)}{1 + d(x_1,x)},$$

which converges to $1$ as $d(x_1, x) \rightarrow \infty$. Here, $C =1+ d(x_0, x_1)$.

Levi
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    @infinity No, it converges to $1$. With the $p$-th power it gets annoying: You have to distinguish a few cases ($p<1$ vs. $p \geq 1$) and pay attention that $(a+b)^p \leq a^p + b^p$ does not always hold. I don't think I want to do this rigorously, sorry. – Levi Nov 09 '19 at 19:50