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Is it possible to define a Cauchy sequence as follows?

Let $(X,d)$ be a metric space and $(x_{n})_{n\in \mathbb{N}}$ be a sequence in it. Then $(x_{n})_{n\in \mathbb{N}}$ is Cauchy iff $\lim_{(j,k)\to (\infty,\infty) }d(x_{j},x_{k})=0$.

Thanks.

Note: By $\lim_{(j,k)\to (\infty,\infty) }d(x_{j},x_{k})=0$, I mean the standard definition of the limit for a function $d:\mathbb{R}\times \mathbb{R} \to \mathbb{R}$ using Euclidean metric $ d_{E}$.

Note2: Thanks for the answers. Conclusion: It can be defined as a double limit indeed, as indicated by http://books.google.co.uk/books?id=lZU0CAH4RccC&pg=PA34&redir_esc=y#v=onepage&q&f=false page 33.

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What precisely is meant by $\lim_{(x,y)\to(\infty,\infty)}f(x,y)=L$, here? –  Cameron Buie May 13 '12 at 21:50
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Yes, assuming that your definition of $\lim\limits_{(j,k)\to(\infty,\infty)}f(j,k)=L$ is that for each $\epsilon>0$ there is $n_0\in\Bbb N$ such that $|f(j,k)-L|<\epsilon$ whenever $j,k\ge n_0$. –  Brian M. Scott May 13 '12 at 21:51
    
Thanks, that's what I meant. –  firemind May 14 '12 at 3:26

2 Answers 2

You have to define properly the double limit $(j,k) \to (\infty, \infty)$. We can take $j=k \to \infty$ and clearly $d(x_j,x_k)=d(x_j,x_j)=0$ but $(x_n)_{n \in \mathbb{N}}$ doesn't necessarily converges (take $x_{2n} = 3 , x_{2n+1} = 5$ for example).

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If you look carefully you need to define what the limit means, is well known that the distance is a continuous map while endowing space $X\times X$ with a certain topology. In this case all its ok and your definition it is equivalent.

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