0

Calculate the diameter of space $ \left( \mathbb{R} , d \right) $ , where $ d : \mathbb{R} × \mathbb{R} \to \mathbb{R} $ is defined $ d(x,y) = \left|\left(\frac{x}{1+ \sqrt{ 1 + x^2}} - \frac{y}{1+ \sqrt{1 + y^2}}\right)\right| $ I don't even know how to start.

amWhy
  • 209,954
user560461
  • 1,735
  • 8
  • 16
  • You have a left parenthesis without a matching right. Do you mean by \abs to be the absolute value of the ensuing expression? – amWhy Nov 18 '18 at 20:48
  • Thanks, i have edited it. Is it clearer now? – user560461 Nov 18 '18 at 20:54
  • Is this what you mean, user560461? Any argument under the square root sign, e.g. in $\sqrt{1+x^2}$, we surround in braces, as follows: \sqrt{1 + x^2} – amWhy Nov 18 '18 at 21:01
  • See if you can find a definition of the diameter of a set in your notes/coursebook. When in doubt, always go back to the definitions. – Rost Nov 18 '18 at 21:08

1 Answers1

1

Hint: Note that \begin{align} \left|\frac{x}{1+\sqrt{1+x^2}}-\frac{y}{1+\sqrt{1+y^2}} \right| \leq \frac{|x|}{1+\sqrt{1+x^2}}+\frac{|y|}{1+\sqrt{1+y^2}} \leq 2 \end{align}

Jacky Chong
  • 25,739