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Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and let $M = X \times Y$. Define $\xi : M \times M \to \mathbb{R}$ by $$\xi((x_1, y_1), (x_2,y_2)) = d(x_1,y_1) + \rho(x_2,y_2).$$ Prove that $(M, \xi)$ is a metric space.

The positivity and symmetry of $\xi$ follow trivially from the fact that $\xi$ is the composition of two metrics. To show that $\xi$ satisfies the triangle inequality, consider that \begin{eqnarray*} \xi((x_1,y_1), (x_2,y_2)) &=& d(x_1, y_1) + \rho(x_2,y_2) \\ & \leq & d(x_1,z_1) + d(z_1,y_1) + \rho(x_2,z_2) + \rho(z_2,y_2) \\ &=& \xi((x_1,z_1),(x_2,z_2)) + \xi((z_1, y_1),(z_2,y_2)). \end{eqnarray*} Prove that $M$ is complete.

Let $(a_n)$ be a Cauchy sequence in $M$. Then $\forall \epsilon > 0 \ \exists N \in \mathbb{N}$ such that $$m,n > N \implies \xi(a_n, a_m) < \epsilon.$$

This problem is really confusing with the fact that we have 4 inputs, I'm struggling to show that $(M, \xi)$ is a complete space.

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    $\begingroup$ There is no additional condition on $X$ and $Y$ to insist $M$ to be complete. What is your condition? $\endgroup$ – Will Kwon Jun 12 '16 at 5:36
  • $\begingroup$ I think you are confusing the metric part. You should reexamine the definition of metric. $y_1$ is in $Y$, not $X$. $\endgroup$ – Will Kwon Jun 12 '16 at 5:47
  • $\begingroup$ Perhaps you mean to write that $\xi((x_1,y_1),(x_2,y_2))=d(x_1,x_2)+\rho(y_1,y_2)$? $\endgroup$ – rnrstopstraffic Jun 12 '16 at 6:23
  • $\begingroup$ No, the question is exactly how I've stated it. Multiple sources state it in this way also. $\endgroup$ – user319128 Jun 12 '16 at 9:32
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HINT: First, as others have pointed out in the comments, your definition of $\xi$ is clearly not correct: $x_1\in X$ and $y_1\in Y$, so $d(x_1,y_1)$ makes no sense, no matter how many sources you may have found for this version. The definition should read

$$\xi(\langle x_1,y_1\rangle,\langle x_2,y_2\rangle)=d(x_1,x_2)+\rho(y_1,y_2)\;.$$

Next, $\langle M,\xi\rangle$ is not necessarily complete. Suppose that $\langle X,d\rangle$ is not complete, and let $\langle x_n:n\in\Bbb N\rangle$ be a $d$-Cauchy sequence in $X$ that has no limit in $X$. Let $y\in Y$ be arbitrary, and for $n\in\Bbb N$ let $p_n=\langle x_n,y\rangle$.

  • Show that $\langle p_n:n\in\Bbb N\rangle$ is $\xi$-Cauchy.
  • Show that $\langle p_n:n\in\Bbb N\rangle$ has no limit in $M$, and conclude that $M$ is not complete.

Thus, if $\langle M,\xi\rangle$ is complete, $\langle X,d\rangle$ has to be complete as well. In similar fashion you can show that if $\langle M,\xi\rangle$ is complete, then $\langle Y,\rho\rangle$ is also complete. Thus, there is no hope of proving that $\langle M,\xi\rangle$ is complete unless you assume that $\langle X,d\rangle$ and $\langle Y,\rho\rangle$ are both complete. The intended exercise was probably to show that if $\langle X,d\rangle$ and $\langle Y,\rho\rangle$ are both complete, then so is $\langle M,\xi\rangle$.

To do this, suppose that $\langle p_n:n\in\Bbb N\rangle$ is a $\xi$-Cauchy sequence in $M$. For $n\in\Bbb N$ let $p_n=\langle x_n,y_n\rangle$.

  • Show that $\langle x_n:n\in\Bbb N\rangle$ is $d$-Cauchy in $X$.
  • Show that $\langle y_n:n\in\Bbb N\rangle$ is $\rho$-Cauchy in $Y$.
  • Conclude that there are $x\in X$ and $y\in Y$ such that $\langle x_n:n\in\Bbb N\rangle\to x$ and $\langle y_n:n\in\Bbb N\rangle\to y$.
  • Show that $\langle p_n:n\in\Bbb N\rangle\to\langle x,y\rangle$.
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