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which of the following metric spaces are complete?

I have doubt to this problem., $X=(0,\pi/2)$ and the metric is $d(x,y)=|\tan x-\tan y|$ is it complete metric space? I took one example $x_n=\frac{1}{2^n}$ is a cauchy sequence in $X$, $d(x_n,x_m)=|\tan\frac{1}{2^n}-\tan\frac{1}{2^m}|\rightarrow 0\notin X$, so it is not complete?

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marked as duplicate by Nate Eldredge, DonAntonio, Arkamis, Hagen von Eitzen, Henry T. Horton Oct 25 '12 at 17:18

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"The space is complete if every Cauchy sequence $x_n$ converges to a point $x$ and this limit point has to be in the space (in your case the space is $X=(0,\frac{\pi}{2})$).

In your case, you picked up the sequence $\left\{\frac{1}{2^n}\right\}$ and you proved that it is a Cauchy sequence. However the limit of this sequence is $0$ and it does not belong to the space $X=(0,\frac{\pi}{2})$. Then, the space is not complete. I hope this is clear.

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What you've done is show that $x_{n}$ is a Cauchy sequence - the distance between $x_{n}$ and $x_{m}$ tends to zero as $n,m\rightarrow\infty$. Now you have to see if it converges to a limit in $X$, which is just as simple. Clearly, $2^{-n}\rightarrow 0$, and so you have a Cauchy sequence in $X$ that doesn't converge in $X$. Thus, $X$ isn't complete.

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I don't see how $2^{-n}$ can "clearly" converge to 0, as 0 isn't in the space and the metric isn't defined at 0. –  Najib Idrissi Oct 25 '12 at 13:39
    
Maybe that wasn't the best way to explain it. What I should have said was that if it were to converge to a limit, that limit would have to be zero. But it can't converge to that, precisely because $0$ isn't in $X$. –  user123123 Oct 25 '12 at 13:42

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