Completion product is product of completions

Question

Let $X,Y$ be metric spaces and $\tilde{X}, \tilde{Y}$ their completions. Is it true that $\tilde{X} \times \tilde{Y}$ is the completion of $X \times Y$? Here both these products have the product metric/topology. I guess this can be proven using the universal property of the completion?

Answer 1

Indeed it can. Since $\overline X\times\overline Y$ is complete and since the natural inclusion from $X\times Y$ into $\overline X\times\overline Y$ is distance-preserving and its range is dense, $\overline X\times\overline Y$ is indeed a completion of $X\times Y$. That's so because, if $Z$ is a complete metric space and $f\colon X\times Y\longrightarrow Z$ is uniformly continuous, $f$ can be extended to one and only one continuous function $F\colon\overline X\times\overline Y\longrightarrow Z$: if $(x,y)\in\overline X\times\overline Y$, you take a sequence $(x_n)_{n\in\Bbb N}$ of elements of $X$ ande a sequence $(y_n)_{n\in\Bbb N}$ of elements of $Y$ such that $\lim_{n\to\infty}x_n=x$ and $\lim_{n\to\infty}y_n=y$ and you define$$F(x,y)=\lim_{n\to\infty}f(x_n,y_n).$$