I need to determine if $((0, 1), d)$ where $d(x, y) = |x^2 - y^2| \forall_{x,y}\in (0,1)$

My argument is as follows, take the sequence defined by $\dfrac{1}{n}$, then we know by $d(x, y)$ that $\dfrac{1}{n} \to 0$ as $n \to \infty$. As $0 \notin (0, 1)$ and the sequence $\dfrac{1}{n}$ is Cauchy w.r.t $d(x, y)$ we have $((0, 1), d) is not a complete metric space by definition.

Also, is $\left(\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right), d \right)$ where $d(x, y) = |tan(x) - tan(y)|$ for all $x, y \in \left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$

Can I just consider $\dfrac{\pi}{2} - \dfrac{1}{n}$? Then I know this sequence converges to $\dfrac{\pi}{2}$ which is not in $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$.

  1. Could somebody highlight how the metric $d(x, y)$ influences the problem?
  • $\begingroup$ "$\tfrac 1n \to 0$" doesn't make any sense when your space is $(0,1)$ regardless of the metric on it. $\endgroup$ – kahen Apr 18 '13 at 0:38
  • $\begingroup$ Are you sure the second sequence is Cauchy in the metric you've defined? $\endgroup$ – Clayton Apr 18 '13 at 0:48
  • $\begingroup$ Could you both clarify further? I feel like I am missing something essential here. $\endgroup$ – CodeKingPlusPlus Apr 18 '13 at 0:55
  • $\begingroup$ $d((\frac{\pi}{2} - \frac{1}{n}), \frac{\pi}{2})$ is not defined in your second metric. $\endgroup$ – John Douma Apr 18 '13 at 3:20

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