### The set of all polynomials

Posted:

**Wed May 17, 2017 2:12 pm**Is the set of all polynomials open in \(\displaystyle{\left(C([-1,1])\,,||\cdot||_{\infty}\right)}\) ?

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Posted: **Wed May 17, 2017 2:12 pm**

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

Posted: **Thu May 18, 2017 7:44 pm**

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).

(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).

Posted: **Thu May 18, 2017 8:47 pm**

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 \(\displaystyle{x=0}\).

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 \(\displaystyle{x=0}\).