If $d(x,y)$ is a metric, then $\frac{d(x,y)}{1 + d(x,y)}$ is also a metric

Let $(X,d)$ be a metric space and for $x,y \in X$ define
$$d_b(x,y) = \dfrac{d(x,y)}{1 + d(x,y)}$$
a) show that $d_b$ is a metric on $X$

Hint: consider the derivative of $f(t)$ = $\dfrac{t}{1+t}$

b) show that $ d$ and $ d_b $ are equivalent metrics.

c) let $(X,d) $ be $(\mathbb{R}, |\cdot|)$ Show that there exists no $ M>0$ such that $ |x-y| $ $\leq$ $Md_b$ $(x,y)$ for all $ x,y$ $\in$ $\mathbb{R}$

I have calculated $f(t)$ and $f'(t)$ and from this I know $f(t)$ is an increasing function as $f'(t)$ is strictly positive. But I don’t know where to go from here or how to do parts b) and c)


a) Separation and symmetry are clear. For the triangular inequality, there is a bit more work.
Let $$f(t):=\frac{t}{1+t}\qquad f'(t)=\frac{1}{(1+t)^2}.$$
Since this function increases on $[0,+\infty)$, the triangular inequality of $d$ yieds
d_b(x,y)=f(d(x,y))\leq f(d(x,z)+d(z,y))=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}
\leq f(d(x,z))+f(d(z,y))=d_b(x,z)+d_b(z,y).

b) You need to show that a sequence converges to $x$ for $d$ if and only if it converges to $x$ for $d_B$.

Assume first that $d(x_n,x)\rightarrow 0$.
d_B(x_n,x)=\frac{d(x_n,x)}{1+d(x_n,x)}\leq d(x_n,x)
so $d_B(x_n,x)$ tends to $0$.

Now if $d_B(x_n,x)$ tends to $0$, $d(x_n,x)$ is bounded by some $M>0$. Indeed, assume for a contradiction that $d(x_n,x)$ is unbounded. So there exists a subsequence $d(x_{n_k},x)$ which tends to $\pm\infty$. Then $d_B(x_{n_k},x)$ must tend to $1$. Contradiction.

\frac{d(x_n,x)}{1+M}\leq \frac{d(x_n,x)}{1+d(x_n,x)}=d_B(x_n,x).
So $d(x_n,x)$ tends to $0$.

c) Observe that $d_B$ is bounded while $|x-y|$ is unbounded. So such a minoration is impossible.

Source : Link , Question Author : Mathsstudent147 , Answer Author : Julien

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