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Let $(X,d)$ be a metric space. Show that both $d_1=\sqrt{d}$ and $d_2=\log(1+d)$ are metrics on $X$.

As far as I know metric is a distance function $d: X\times X\to\mathbb{R}$ that takes two inputs such as $d(x,y)=|x-y|$. But both $d_1$ and $d_2$ have only one input $d$.

And if we are going to show that a function is a metric, do we just need to show that it satisfies:
1. $d(x,y)\geq0$, $d(x,y)=0$ iff $x=y$,
2. $d(x,y)=d(y,x)$,
3. $d(x,y)\leq d(x,z)+d(z,y)$ ?

My dumb guess is that we make:
$(d_1)^2=d$ and $e^{d_2}=1+d$, so $e^{d_2}-(d_1)^2=1$ (so now we have two variables $d_1$ and $d_2$)

But it is still not very clear what should we do with the $d_1$ and $d_2$? In the case of $d(x,y)=|x-y|$ in $\mathbb{R}^n$, we can treat $x$ and $y$ as vectors and use their components. But the $X$ that we have here is a general metric space.

I am confused.

Thanks for the help!

user71346
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  • The third property should be the opposite: $d(x,y)\leq d(x,z)+d(z,y)$ – AugSB Oct 27 '15 at 11:57
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    $d_1 =\sqrt{d}$, where $d$ is a metric. This means $d_1(x,y) = \sqrt{d(x,y)}$, so $d_1$ is a function of two variables. – GEdgar Oct 27 '15 at 12:04

1 Answers1

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Note that you are given a metric space $(X,d)$, where $d$ is a metric. Thus, $d$ satisfies the three properties you have enumerated for every $x,y\in X$. Assuming that $d$ satisfies those properties, now you have to show that both $d_1(x,y)=\sqrt{d(x,y)}$ and $d_2(x,y)=\log(1+d(x,y))$ satisfy them too.

AugSB
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