MHB Problem of the Week #110 - July 7th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Show that if $d$ is a metric for $X$, then \[d^{\prime}(x,y) = \dfrac{d(x,y)}{1+d(x,y)}\]
is a bounded metric that gives the topology of $X$.

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Hint: [sp]If $f(x)=x/(1+x)$ for $x>0$, use the mean value theorem to show that $f(a+b)-f(b)\leq f(a)$.[/sp]

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge. You can find his solution below.

[sp]Since $d$ takes values in $[0, \infty)$, $d'$ takes values in $[0,1)$. In particular, $d'$ is bounded. Now $d'(x, y) = 0$ if, and only if, $d(x, y) = 0$, which occurs if, and only if, $x = y$. Symmetry of $d'$ follows immediately from symmetry of $d$.

To verify the triangle inequality for $d'$, consider the function $f : [0, \infty) \to [0, \infty)$ given by $f(x) := x/(1 + x) = 1 - 1/(1 + x)$. Note (1) $f$ is increasing and (2) $f(u + v) \le f(u) + f(v)$ for all $u, v\in [0, \infty)$. Property (2) follows from the fact that

$f(u + v) - f(u) = \dfrac{v}{(1 + u)(1 + u + v)} \le \dfrac{v}{1 + v} = f(v)$

for all $u, v \in [0, \infty)$. Using (1), (2), and the triangle inequality for $d$, we have, for all $x, y, z \in X$,

$d'(x, z) = f(d(x, z)) \le f(d(x, y) + d(y, z)) \le f(d(x, y)) + f(d(y, z)) = d'(x, y) + d'(y, z)$.

Thus the triangle inequality is satisfied by $d'$, and $d'$ is a metric on $X$.

To finish the proof, it suffices to show that $d'$ is equivalent to $d$. Let $x_n \to x$ in $(X, d)$. Then $d'(x_n, x) \le d(x_n, x) \to 0$ as $n \to \infty$. Thus $x_n \to x$ in $(X, d')$. Conversely, if $x_n \to x$ in $(X, d')$, then for a given $\epsilon \in (0,1)$ there corresponds a positive integer $N$ such that $d'(x_n, x) < e$ for all $n \ge N$, i.e., $d(x_n, x) < \epsilon/(1 - \epsilon)$ for all $n \ge N$. Hence $\varlimsup d(x_n, x) \le \epsilon/(1 - \epsilon)$ for all $\epsilon \in (0,1)$. Taking the limit as $\epsilon \to 0^+$ gives $\varlimsup d(x_n, x) = 0$. Hence $x_n \to x$ in $(X, d)$.[/sp]