Let $(x_n)$ and $(y_n)$ be Cauchy sequences in $(X, d)$. Show that $(x_n)$ and $(y_n)$ converge to the same limit iff $d(x_n, y_n) \rightarrow 0$

Proof $\rightarrow$

Suppose $(x_n) \to a$ and $(y_n) \to a$, then \begin{equation} \forall_{\epsilon > 0} \exists_{N_1} s.t. \forall_{n>N_1} \implies d(x_n, a) < \dfrac{\epsilon}{2} \\ \forall_{\epsilon > 0} \exists_{N_2} s.t. \forall_{n>N_2} \implies d(y_n, a) < \dfrac{\epsilon}{2} \end{equation}

Given $\epsilon>0$ take $N=\max(N_1, N_2)$ then $\forall_{n>N}$ we have: \begin{align} d(x_n, y_n) &\leq d(x_n, a) + d(a, y_n) \\ & < \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} = \epsilon \end{align}

What is the cleanest way to finish this piece of the proof? Technically I feel like I could do the following:

Consider $d(d(x_n, y_n), 0)$: \begin{align} d(d(x_n, y_n), 0) &< d(\epsilon, 0) < \epsilon \implies d(x_n, y_n) \to 0 \end{align}

What can I do to make this direction of the proof cleaner? Note that I only used the fact that the sequences converged and nothing special about Cauchy sequences. Additionally, what is the formal definition of $d(x_n, yn) \to 0$?


$\leftarrow$ Suppose $d(x_n, y_n) \to 0$ that is $\forall_{\epsilon > 0} \exists_{N_1} s.t \forall_{n > N_1} \implies |d(x_n, y_n) - 0| < \epsilon$. Given $\epsilon > 0$ take $N > N_1$ then by the triangle inequality we have:

\begin{align} 0 < d(x_n, y_n) \leq d(x_n, a) + d(a, y_n) < \epsilon \end{align} We know this $a$ exists as $(X, d)$ is a complete metric space and $(x_n)$ and $(y_n)$ are Cauchy. Thus we have that $d(x_n, a) < \epsilon \land d(y_n, a) < \epsilon$. Therefore, $(x_n)\to a \land (y_n)\to a$

How does the other direction look?

  • 1
    $\begingroup$ The reverse implication is false if the space $(X,d)$ is not complete. $\endgroup$ – egreg Apr 17 '13 at 21:31

Remember that your metric d is real valued, and the real numbers are a metric space. Thus, the formal definition of $d(x_n, y_n) \rightarrow 0$ is the same as for convergence of a sequence in a metric space, i.e., for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ we have $d_{\mathbb{R}}(d(x_n,y_n), 0) < \epsilon$.

Note the $d_{\mathbb{R}}$! This is the metric on the real numbers (where $d(x_n,y_n)$ and $0$ live). It is defined by $d_{\mathbb{R}}(x,y) = |x -y|$. So a much less obfuscated way to write the definition above is the following:

$d(x_n, y_n) \rightarrow 0$ if for every $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ we have $|d(x_n,y_n) - 0| < \epsilon$. But this is exactly what you have shown in the part before your proposed finish! You can actually stop right there with the $\epsilon$. You could also add a note about how you know $d(x_n,y_n) \geq 0$.

The error in what you go on to write is that $d(d(x_n,y_n), 0)$ may not be defined since $d$ is a metric on $X$ while $d(x_n, y_n)$ and $0$ are in $\mathbb{R}$.

As for the Cauchy assumption, you need that to do the converse (along with completeness of $X$).

  • $\begingroup$ take a look at the other direction of the proof I just added $\endgroup$ – CodeKingPlusPlus Apr 17 '13 at 23:31

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