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Could someone help me through this problem? Let $X$ be a metric space. Show that if $\{x_{n}\}$ and $\{y_{n}\}$ are Cauchy sequences in $X$ then $d(x_{n},y_{n})$ converges in $\mathbb{R}$.

Does this follow from the fact that every Cauchy sequence in $\mathbb{R}$ is convergent?

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    It does follow from this that Cauchy sequences in $\mathbb{R}$ converge; however, I believe you will have to use that fact to prove this one.2012-04-26

4 Answers 4

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If $X$ is a complete metric space then there exist $ x,y \in X $ such that $ d(x_n,x) \rightarrow 0$ and $ d(y_n,y) \rightarrow 0$. Thus \begin{eqnarray*} \vert d(x_n,y_n) - d(x,y) \vert & = & \vert d(x_n,y_n) - d(x,y_n) + d(x,y_n) - d(x,y)\vert \\ & \le & \vert d(x_n,y_n) - d(x,y_n) \vert + \vert d(x,y_n) - d(x,y)\vert \\ & \le d(x_n,x) + d(y_n,y) \rightarrow 0 \\ \end{eqnarray*} If $X$ isn't complete, I dont believ the result is true because the natural candidate $d(x,y)$ maybe don't exist.

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    You don't need the limit points $x$ and $y$. Look at Davide's answer: the right hand side of the inequality can be estimated from above by $d(y_n,y_m) + d(x_n,x_m)$ which can be made arbitrarily small by choosing $m,n$ large enough.2012-04-26
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Hint: use triangular inequality $|d(x_n,y_n)-d(x_m,y_m)|\leq |d(x_n,y_n)-d(x_n,y_m)|+|d(x_n,y_m)-d(x_m,y_m)|,$ then the reversed triangular inequality $|d(x,y)-d(z,y)|\leq d(x,z)$.

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    @Aditya Yes, it was a typo. Thanks.2012-04-26
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This is essentially Davide Giraudo's approach, but somewhat shorter: The triangle inequality gives $d(x_m,y_m)\leq d(x_m,x_n)+d(x_n,y_n)+d(y_n,y_m)$ or $d(x_m,y_m)-d(x_n,y_n)\leq d(x_m,x_n)+d(y_n,y_m)\ .$ As the right side is symmetric in $m$ and $n$ we have in fact $\bigl|d(x_m,y_m)-d(x_n,y_n)\bigr|\leq d(x_m,x_n)+d(y_n,y_m)\ .$

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Every cauchy sequence in $\mathbb{R}$ is convergent, so exists $x, y\in\mathbb{R}$ such that $x_n\to x$ , $y_n\to y$, then by continuity of the metric, $d(x_n,y_n)\to d(x,y).$

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    $x_n$ and $y_n$ are sequences in $X$, not in $\mathbb{R}$.2012-04-26