# Determining Complete Metric Spaces

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