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Is the following space complete?

$X_1=\left(0,\dfrac{\pi}{2}\right)$ defined by $d (x,y)=|\tan x-\tan y \ |$

Let $x_n$ be a Cauchy sequence in $X$ then, we will have $n,m\in \mathbb N$ such that $d(x_n,x_m)<\epsilon$ for any arbitrary $\epsilon>0$

$\implies |\tan x_n-\tan x_m|<\epsilon$.Please help me to complete this.

Julian Rachman
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2 Answers2

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The space is not complete. Show that the sequence $a_n=\dfrac{1}{n}$ is a Cauchy sequence, which doesn't converge (in the given space-it actually converges to $0$).

voldemort
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Note that $\tan$ is continuously differentiable on $(-\pi/2,\pi/2)$. In particular, for each compact subset $K\subset(-\pi/2,\pi/2)$ there is some $L>0$ such that $|\tan(x)-\tan(y)|\leq L|x-y|$ for each $x,y\in K$. Now, if you choose $K=[-\pi/4,\pi/4]$, then this shows, that any sequence $(x_n)$ which converges to 0 with respect to the Euclidian metric also converges to $\tan(0)=0$ with respect to your metric $d$ and that these sequences are Cauchy-sequences in $X_1$. But then $X_1$ cannot be complete (take your favourite 0-sequence), since $0\notin X_1$.

sranthrop
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  • well I think it's easier than that. Thus note that $tan(a_n)$ converges to $0$, as tan is continuous function in $\mathbb{R}$, and hence the sequence is Cauchy. Here $a_n$ is any sequence converging to $0$. So you do not need that tan is differentiable. But still- this is a good approach, so +1. – voldemort Jan 12 '15 at 05:43
  • Actually, you need that $\tan$ is uniformly continuous on any compactum containing 0, and this is indeed so since continuous functions are uniformly continuous on compact sets. So in this sense you are right. An interesting question would be: Is there a function $f$ such that $d(x,y)=|f(x)-f(y)|$ is a metric, but our argument fails? – sranthrop Jan 12 '15 at 06:03