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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?

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Proxy123
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1 Answers1

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As defined in the link

$$d_{X\times Y}\left((x_n,y_n),(x_m,y_m)\right)=\max\left(d_X(x_n,x_m),d_Y(y_n,y_m)\right)$$

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.

marwalix
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  • And if $ d_{X\times Y} $ is any other metric that induces the product topology? – Proxy123 Dec 22 '16 at 05:58
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    Cauchy property is a metric property i.e depends on the metric, meaning that two metrics inducing the same topology may have different Cauchy sequences. I have to look back far in my memory to show you a counterexample – marwalix Dec 22 '16 at 06:04
  • Yeah, I must have misread the question I was set. Thanks for your help – Proxy123 Dec 22 '16 at 06:14