I came across a proof where it's used one such map in the following way. Let $V \subsetneq \mathbb{R}^{n}$ be an open, non empty, subset - we're using euclidean topology. We consider a projection map $\pi : V \rightarrow \mathbb{R}$ such that $\forall x \in V \ \pi(x) \geq 0$.
I just cannot convince myself that one such map always exists. Could anyone please explain why? I would also appreciate an example.
Thanks a lot.
On
There is always the trivial affine map $\pi(v)=0$ which works for all $V$. But if we restrict to rank one maps, we can say the following:
Theorem. Given $V\subset\Bbb R^n$. There exists a rank one affine map $\pi:\Bbb R^n\to \Bbb R$ with $\pi(v)\ge 0$ if and only if $V$ is contained in some half-space.
Proof.
The rank one affine maps $\pi:\Bbb R^n\to\Bbb R$ are exactly the maps $\pi(v)=\langle c,v\rangle-\lambda$ for some non-zero $c\in\Bbb R^n$ and some $\lambda\in\Bbb R$. The statement $\pi(v)\ge 0$ is then equivalent to
$$\langle c,v\rangle \ge \lambda,$$
which defines the desired half space.
This means that such a "projection" exists if e.g. $V$ is bounded or a proper affine subspace. It does not work e.g. for complements of bounded sets.
If $n=1$ and $V = \mathbb{R}$, then no such map exists. (Unless projection map doesn't mean what I think it means).