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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 $\mathbb{R}$.

Is this it follows that every Cauchy sequence in $\mathbb{R}$ is convergent?

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Is the metric space complete? – TenaliRaman Apr 25 '12 at 20:50
@TenaliRaman You don't need the metric space to be complete, since the convergence in question is in $\mathbb{R}$ – Brett Frankel Apr 25 '12 at 20:54
@BrettFrankel I just remember proving this for a complete metric space at some point, basically I had assumed $\{y_n\} \rightarrow y$ and $\{x_n\} \rightarrow x$ and used this to show $\{\langle x_n, y_n \rangle\} \rightarrow \langle x, y \rangle$. Then, showing $d(x_n, y_n)$ converges was quite easy. I have to check Davide's hint below to see how the same can be without assuming completeness. – TenaliRaman Apr 25 '12 at 20:59
@TenaliRaman: You can fix that proof by running it in the completion of $X$. – Chris Eagle Apr 25 '12 at 22:30
@breton This is problem 23 in chapter 3 of baby rudin. It gives a hint that is basically the ones below. In your post, $d(x_n,y_n)$ is a sequence, so it should be written $\{d(x_n,y_n)\}$ or similarly. – john w. Apr 26 '12 at 13:44
up vote 3 down vote accepted

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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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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In the reverse triangle inequality, the RHS should be $d(x,z)$. – Aditya Apr 26 '12 at 1:56
@Aditya Yes, it was a typo. Thanks. – Davide Giraudo Apr 26 '12 at 9:06

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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I can not still use the continuity of the metric – Breton Apr 25 '12 at 23:48
$x_n$ and $y_n$ are sequences in $X$, not in $\mathbb{R}$. – Chris Eagle Apr 26 '12 at 10:09

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. – t.b. Apr 26 '12 at 0:42

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