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Let $(V, \|\cdot \|_0)$ and $(W, \|\cdot \|_1)$ be normed vector spaces.
Given that $(\overline{V},\overline{d}_0)$ is a completion of $(V,d_0)$ (where $d_0(v,x)=\|v-x\|_0$), and $(\overline{W},\overline{d}_1)$ is a completion of $(W,d_1)$ (with $d_1(y,w)=\|y-w\|_1$), then $(\overline{V}\times \overline{W},\overline{d})$ is a completion of $(V\times W, d)$ with $\overline{d}((\overline{x}, \overline{y}),(\overline{v}, \overline{w}))=\overline{d}_0(\overline{x}, \overline{v})+\overline{d}_1(\overline{y}, \overline{w})$, where $d((x,y),(v,w))=\vertiii{(x,y)-(v,w)}=\|x-v\|_0+\|y-w\|_1$.
As far as I understand, the argument should be based on the fact that, since $(\overline{V}, \overline{d}_0)$ is a completion of $(V,d)$, there is an isometry $\overline{\iota}:V\to \overline{V}$ such that $\overline{\overline{\iota}(V)}=\overline{V}$ and $d_0(\overline{\iota}(x),\overline{\iota}(v))=d_0(x,v)$. Similarly, there is an isometry $\overline{\phi}:W\to \overline{W}$.
Now, I'm not quite sure what to do next, since I'm not convinced that I fully understand how these isometries work, so just want to make sure I'm doing it right. Here's how my proof goes next:
Thus $(\overline{\overline{\iota}(V)}\times \overline{\overline{\phi}(W)})=\overline{V}\times \overline{W}$ with $\overline{d_0}(\overline{\iota}(x),\overline{\iota}(v))= \overline{d_0(x,v)}=\overline{d}_0(\overline{x},\overline{v})$ and $\overline{d_1(\overline{\phi}(y),\overline{\phi}(w))}= \overline{d_1(y,w)}=\overline{d}_1(\overline{y},\overline{w})$. Thus $\overline{d}((\overline{x},\overline{v}),(\overline{y},\overline{w}))=\overline{d}_0(\overline{v},\overline{x})+\overline{d}_1(\overline{w},\overline{y})$.
Please let me know if I'm not missing something in my argument.
Also, what is the difference between a completion of a metric space and its closure? Is it that $\mathbb{R}$ is a closure of $\mathbb{Q}$, but it is also a completion of $\mathbb{R}$, while $\mathbb{C}$ is a completion but not a closure of $\mathbb{Q}$?