# Is the set of strictly positive polynomials an open set? In case it is a smooth manifold.

Let $$I=[-1, 1]$$, $$\mathcal{P}^n$$ be the set of polynomials of degree $$n$$ with domain $$I$$, $$\mathcal{P}^n_+(I, \mathbb{R})$$ be the set of real-valued polynomials with domain $$I$$ that are strictly positive $$\forall p\in\mathcal{P}^n_+(I, \mathbb{R}) \ \ p(t) >0 \forall t\in I$$ Is this set open with respect to the topology induced by the $$C^0$$ norm, i.e. $$\| p \|_{C^0} = \sup_{t\in I} | p(t) |$$

Attempt of proof: In order to prove that $$\mathcal{P}_+(I, \mathbb{R})$$ is open with respect to the mentioned topology I will try to demonstrate that all its elements has a neighborhood contained in it.

1. First we define a ball as $$N(p, r) = \left\{q\in\mathcal{P}^n\text{ s.t. } \| p - q \|_{C^0}
2. Any ball is a neighborhood.
3. We desire to prove that $$p\in \mathcal{P}_+^n(I, \mathbb{R}) \implies \exists r>0\text{ s.t. } N(p, r) \subset \mathcal{P}^n_+(I, \mathbb{R})$$ Let $$p\in\mathcal{P}_+^n$$ and the ball $$N'(p, r) = \left\{p+e\in\mathcal{P}^n\text{ s.t. } \| e \|_{C^0}

we can always find a sufficiently small $$r>0$$ such that $$\inf_{t\in I} \left\{ p(t) + e(t)\right\} >0$$

It this proof right? If it is, then as $$\mathcal{P}^n$$ is a vector space, then $$\mathcal{P}_+^n$$ is a smooth manifold.

You haven't said how you can find such an $$r$$. We need to use something like the compactness of $$I$$.
To elaborate: Let $$p \in \mathcal{P}^{+}_{n}(I, \mathbb{R})$$. Then, since $$p(t) > 0$$ for all $$t \in I$$, we see that $$\delta := \frac12\inf_{t \in I} p(t)$$ is positive (since the $$\inf$$ is actually attained).
Now, given any $$q \in \mathcal{P}_{n}$$ in the $$\delta$$-neighbourhood of $$p$$ and $$t \in I$$, we have $$q(t) = q(t) - p(t) + p(t) \geqslant p(t) - |q(t) - p(t)| \geqslant p(t) - \delta > 0.$$
Note that compactness is needed. For example, if we change $$I$$ to $$(0, 1)$$, then you cannot find any such neighbourhood around the polynomial $$p(t) = t$$.