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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 $$ N(p, r) = \left\{q\in\mathcal{P}^n\text{ s.t. } \| p - q \|_{C^0}<r\right\} $$
  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 $$ N'(p, r) = \left\{p+e\in\mathcal{P}^n\text{ s.t. } \| e \|_{C^0}<r\right\} $$

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.

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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 $p(t) = t$.

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