# Cartesian Product of Two Complete Metric Spaces is Complete

This is a problem I'm stuck on that our professor gave us for additional practice (not homework, but its recommended that we understand how to prove it).

We know X and Y are complete metric spaces, and we need to show that $X \times Y$ is complete. I'm really lost on the proof technique. We were given an outline as follows, but I could only fully figure out (1). Part 3 is what we've been really stuck on though. I was wondering whether someone could give an proof for say a more specific space where $X = \mathbb{R}, Y = \mathbb{R}$, so I could understand the principle.

Outlined:

1) Show that $d_{X \times Y} ( (a_1,b_1) , (a_2,b_2)) = \max \{ d_X (a_1,a_2) , d_Y (b_1, b_2)\}$ is a metric.

2) Prove that this gives the product topology on $X \times Y$.

3) Prove that if $a_n, b_n$ are Cauchy sequences, where $a_n \in X$ and $b_n \in Y$, then $(a_n,b_n )$ is Cauchy.

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Thanks a lot for all the clarifications. They are really helping me get a better picture of this sort of problem, in particular the sort that involves Cartesian products of two spaces. – Rishi Oct 31 '10 at 23:09
1) is a little strange. You can show that $d_{X\times Y}$ is a metric. That it is the product metric, is not a theorem, but a definition. Moreover it is not the only reasonable definition. Note, e.g., that for $X=Y=\mathbb{R}$,$d_{X\times Y}$ is not the "natural" Euclidean distance. – Stefan Walter Nov 17 '10 at 14:14

As for (2), it suffices to prove that the balls produced by the distance $d_{X\times Y}$ are a basis for the product topology.

So let's write down what it is a ball for the distance $d_{X\times Y}$ with center $(a,b)$ and radius $\varepsilon > 0$:

$$B_{X\times Y}((a,b); \varepsilon) = \left\{ (x,y) \in X\times Y \ \vert \ d_{X\times Y} ((a,b), (x,y))= \max \left\{ d_X(a,x) , d_Y (b,y) \right\} < \varepsilon \right\} \ .$$

But

$$\max \left\{ d_X(a,x) , d_Y (b,y) \right\} < \varepsilon \quad \Longleftrightarrow \quad d_X(a,x) <\varepsilon \ \text{and} \ d_Y(b,y) < \varepsilon \ .$$

Hence we see that

$$B_{X\times Y }((a,b); \varepsilon)\ = \ B_X (a;\varepsilon ) \times B_Y (b; \varepsilon ) \ .$$

That is: a ball for the distance $d_{X \times Y}$ is the same as a product of balls for the distances $d_X$ and $d_Y$. Which means that balls for the distance $d_{X\times Y}$ form a basis for the product topology.

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For 3) if you are given an $\epsilon$, you can find $N$ such that $|a_n-a_m|<\epsilon \text{ if } n,m>N$ and similarly for $M$ and the $b$ series. Doesn't $\max{(N,M)}$ give a limit that $|(a_n,b_n)-(a_m,b_m)|<\epsilon$?

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If ${a_n}$, ${b_n}$ are Cauchy, then for some N, $d_X(a_n,a_m) < \epsilon$, $d_Y(b_n,b_m) < \epsilon$ when $n,m \geq N$. Because of this, $d_{X \times Y} ((a_n,b_n),(a_m,b_m)) < \epsilon$ for $n,m \geq N$.

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Actually if we want to prove that a product space is complete then we have to take a Cauchy sequence from the product space and then show that it converges to a point in it.

Observe that for all $a_1, a_2 \in X$ and $b_1, b_2 \in Y,$$d_X (a_1,a_2) \leq \max \{ d_X (a_1,a_2) , d_Y (b_1, b_2)\}\tag1$ $d_Y (b_1,b_2) \leq \max \{ d_X (a_1,a_2) , d_Y (b_1, b_2)\} \tag2$.

Suppose we have a Cauchy sequence $((a_n, b_n))$ in $X \times Y$. Then given $\epsilon > 0$ there exist $N$ such that for all $n,m \geq N$ we have $\max \{ d_X (a_n,a_m) , d_Y (b_n, b_m)\} < \epsilon$

Due to $(1)$ and $(2)$ we have $d_X (a_n,a_m) < \epsilon$ and $d_Y (b_n,b_m) < \epsilon$ for all $n,m \geq N$ implying that $(a_n)$ and $(b_n)$ are Cauchy in $X$ and $Y$ respectively. Since $X$ and $Y$ are both complete $(a_n)$ converges to some $a \in X$ and $(b_n)$ converges to some $b \in Y$ and thus $(a,b) \in X \times Y.$

Now it is a matter of showing that $((a_n, b_n))$ converges to $(a,b)$ in $X \times Y$ which is easy to do since $\lim_{n \to \infty }d_{X \times Y} ( (a_n,b_n) , (a,b)) = \lim_{n \to \infty } \max \{ d_X (a_n,a) , d_Y (b_n, b)\} = 0.$

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