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.
Best Answer
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.