Embedding of an open subset

Question

How can one embedd an open subset $V$ of $\mathbb R^n$ into $\mathbb R^{n+1}$ with a closed map? Can one make that embedding smooth? I had some idea like $F \colon V \to \mathbb R^{n+1}$ given by $F(x)=(x,dist(x,\partial V))$. But I don't know how to show if the map is closed.

Answer 1

I won't address the smoothness issue, but following your suggestion (with altered notation) I'll use the function $$x \mapsto \left(x,\frac{1}{d(x,\mathbb{R}^n-V)}\right) $$ where $d(x,\mathbb{R}^n-V)$ denotes the infimum of $d(x,y)$ over all $y \in \mathbb{R}^n-V$.

The fact that this map is closed follows by applying continuity of the function $x \mapsto d(x,\mathbb{R}^n-V)$. To see this, consider a sequence $(x_i,\frac{1}{d(x_i,\mathbb{R}^n-V)})$ in the image that converges to a limit $(x,a) \in \mathbb{R}^{n+1}$. The sequence $x_i$ in the domain $V$ converges to $x \in \mathbb{R}^n$. If $x \in V$ then by applying continuity we get $d(x,\mathbb{R}-V)=\frac{1}{a}$ and you're done. If $x \in \mathbb{R}-V$ then $d(x_i,\mathbb{R}^n-V)$ converges to zero, but this contradicts that $\frac{1}{d(x_i,\mathbb{R}^n-V)}$ converges to $a$.