# Regarding Weierstrass approximation theorem and simply connected domains

Let $$G$$ be an open connected set in $$\mathbb{C}^n$$. Let $$u,v\in G$$. Now we know that their exists a path $$\gamma:[0,1]\longrightarrow G$$ Such that $$\gamma(0)=u$$ and $$\gamma(1)=v$$. Then it is said that by the Weierstrass approximation theorem, we find a polynomial $$P:[0,1]\longrightarrow G$$ such that $$P(0)=u$$ and $$P(v)=v$$. Then it is said that it is easy to choose a simply connected domain $$D\in \mathbb{C}, [0,1]\subset D,$$ such that $$P(\lambda)\in G$$ for every $$\lambda\in D$$.

My questions are;

1) How do we get such a $$P$$. I know that $$\gamma$$ can be approximated uniformally by polynomials $$P_n$$. But how do we get a $$P$$ among them such that $$P(0)=u$$ and $$P(v)=v$$. Like the sequence $$\frac{1}{n}\longrightarrow 0$$. But no member of the sequence is equal to zero.

2)What theorem guarantees the existence of such a simply connected domain $$D$$?

Suppose $$\|\gamma-P\|_\infty < \epsilon$$ and let $$l(t) = (1-t)(\gamma(0)-P(0)) + t (\gamma(1)-P(1))$$ and let $$Q=P+l$$. Then $$\|\gamma-Q\|_\infty \le \|\gamma-P\|_\infty+ \|l\|_\infty < 2 \epsilon$$.
Since $$G$$ is open, for sufficiently small $$\epsilon>0$$ we have $$Q(t) \in G$$ and $$Q(t) = \gamma(t)$$ for $$t \in \{0,1\}$$.
Suppose you have a polynomial $$P$$ such that $$P([0,1]) \subset G$$. This is smooth and defined everywhere on $$\mathbb{C}$$. In particular, $$P^{-1}(G)$$ is open and contains $$[0,1]$$. In particular, we can find some $$\delta>0$$ such that $$[0,1]+B(0,\delta) \subset P^{-1}(G)$$. This is a suitable simply connected domain.