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## The set of all polynomials

Functional Analysis
Papapetros Vaggelis
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### The set of all polynomials

Is the set of all polynomials open in $\displaystyle{\left(C([-1,1])\,,||\cdot||_{\infty}\right)}$ ?
r9m
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### Re: The set of all polynomials

Polynomials being a subspace of Banach Space $(C[-1,1],\lVert \cdot \rVert_{\infty})$, must have empty interior, otherwise it'd have to be the full space.

(One argument for showing it's not closed, Polynomials have a countable Hamel basis and Banach spaces of infinite dimension cannot have a countable Hamel basis).
Papapetros Vaggelis
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### Re: The set of all polynomials

Thank you r9m.

Here is another idea.

Suppose that $\displaystyle{\mathcal{P}}$ (the set of polynomials) is open in $\displaystyle{\left(C([-1,1]),||\cdot||_{\infty}\right)}$.

Let $\displaystyle{f(x)=x\,,x\in\left[-1,1\right]}$ and then $\displaystyle{f\in\mathcal{P}\subseteq C([-1,1])}$.

There exists $\displaystyle{\epsilon>0}$ such that $\displaystyle{B(f,\epsilon)\subseteq \mathcal{P}}$.

Consider the continuous function $\displaystyle{g(x)=f(x)+\dfrac{\epsilon\,x}{2\,(1+|x|)}\,,x\in\left[-1,1\right]}$

Observe that $\displaystyle{\forall\,x\in\left[-1,1\right]\,\,,|g(x)-f(x)|=\dfrac{\epsilon\,|x|}{2\,(1+|x|)}<\dfrac{\epsilon}{2}}$

so, $\displaystyle{||g-f||_{\infty}\leq \dfrac{\epsilon}{2}<\epsilon\implies g\in B(f,\epsilon)\implies g\in\mathcal{P}}$

which is a contradiction since $\displaystyle{g}$ is not twice differentiable at $\displaystyle{x=0}$.