# Cauchy sequences product metric

I've been trying to prove that if $x_n, y_n$ are Cauchy then so is $(x_n, y_n)$ when X x Y has a metric that induces the product of the metric topologies on X and Y, and apparently I'm missing something quite obvious because, referring to

Cartesian Product of Two Complete Metric Spaces is Complete

it is apparently as simple as saying that since we have an N for which $d_X (x_n, x_m) < \epsilon$ and an M for which $d_Y(y_n, y_m) < \epsilon$ then choose the maximum of these, giving $d_{X\times Y}((x_n,y_n), (x_m, y_m)) < \epsilon$

But I don't understand the logical step. What do we know about the metric of the product that makes this true?

• There are many ways to define a metric on a cartesian product of metric spaces. – user333870 Dec 22 '16 at 4:16
• I am aware of this.. – Proxy123 Dec 22 '16 at 4:17
• Is your question referring to a specific metric on the product space then? – user333870 Dec 22 '16 at 4:18
• Any metric that induces the product of the metric topologies on X and Y – Proxy123 Dec 22 '16 at 4:21
• $\max \{ d_X , d_Y \}$ induces the product topology on $X \times Y$. – Theoretical Economist Dec 22 '16 at 4:25

$$d_{X\times Y}\left((x_n,y_n),(x_m,y_m)\right)=\max\left(d_X(x_n,x_m),d_Y(y_n,y_m)\right)$$
And so when we take the largest of $N$ and $M$ both $d_X(x_n,x_m)$ and $d_Y(y_n,y_m)$ are smaller than $\epsilon$ for all $n,m\geq \max(N,M)$. And so is their maximum i.e the product distance.
• And if $d_{X\times Y}$ is any other metric that induces the product topology? – Proxy123 Dec 22 '16 at 5:58