Consider $a_{i},\; i=1,...,n$, real distinct numbers and the polynomial of degree odd $n$, $$p(x) = \prod_{1}^{n}(x-a_{i}).$$ Define $$A = \{b\in\mathbb{R} \mid p(x) = b\text{ has $n$ distinct roots}\}.$$ (i) Show that $0 \in A$. (ii) Show that $A$ is bounded. (iii) Use the Implicit Function Theorem to show that $A$ is open.
(i) is trivial
(ii) I take a sequence $(b_{i})$ such that $b_{i} \to 0$ to get a contradiction, but I couldn't. I just wanted a hint, some better way.
(iii) I never solved any questions involving IFT in order to classify a set. I have no idea how to use it. I wish someone could help me with that. I imagine that I should define $f: \mathbb{R}^{p} \to \mathbb{R}^{q}$ ($p \leq q$) such that $A \subset \mathbb{R}^{p}$ and take $x_{0} \in A$ and to use the IFT to get $\varphi: B_{\delta}(x_{0}) \subset A \to \mathbb{R}^{q} \in C^{1}$. It's the only way I can imagine using IFT. Anyway, I appreciate any help!
You can actually also apply the IFT $n$ times on $p:\mathbb{R}^1 \rightarrow \mathbb{R}^1$ as follows:
Proof of (iii): Let $b_0 \in A$. then $p(x)=b_0$ has $n$ distinct roots $x_{0,i}$. As $p$ is of degree exactly $n$, the roots must all be of multiplicity 1 and therefore $p'(x_i) \neq 0$. Hence, the IFT can be applied $n$ times to $p(x)$ at the positions $x_{0,i}$. This means we get continuous functions
$$ x_i= x_i(b), $$
defined for $\|b-b_0\|$ small enough, such that
$$p(x_i(b)) = b$$ and $$x_i(b_0) = x_{0,i}.$$
That means that for small variations in $b$ around $b_0$ the roots of $p(x)=b$ vary also small and are still distinct if the variation is small enough. (I'm too lazy to carry out the epsilon delta formulations here ... :) ) Hence $A$ contains a neighborhood around $b_0$ and is open.