Study is the set of polynomial functions is closed or open in $(C[-1,1], \|\cdot\|_\infty)$? I think that this set cannot be open, however I don't know how to proof that it is closed. Could someone help me please?.
Thanks.
 A: Let $A$ be the set of polynomials in $C[-1,1]$.
Then $A$ is not closed because of the sequential characterization of closedness.
We can find a sequence of polynomials which converges to a function $f$ which is not a polynomial because $A$ is dense in $(C[-1,1],||.||_{\infty})$ 
Also is not open because its complement,namely $A^c$, is not closed.
Again from sequential characterization of closedness,take the sequence $f_n \in A^c$ such that $f_n(x)=\frac{1}{n(x+2)} \to^{||.||_{\infty}}0 \notin A^c$
A: The polynomials are not closed in $C[-1,1]$ with the max-norm, because it is well known that they are dense in it, and not every continuous function is a polynomial.
A: It is known that $(C[-1,1], \|\cdot\|_\infty)$ is a Banach space.
If the space of polynomials $\mathcal{P}[-1,1]$ were a closed subset of $C[-1,1]$, then $(\mathcal{P}[-1,1], \|\cdot\|_\infty)$ would also be a Banach space.
The countable set $\{1, x, x^2, \ldots \}$ is a Hamel basis for $\mathcal{P}[-1,1]$.
However, infinite-dimensional Banach spaces necessarily have uncountable Hamel bases, as you can see here.
This is a contradiction, hence $\mathcal{P}[-1,1]$ cannot be closed.
$\mathcal{P}[-1,1]$ also cannot be open in $C[-1,1]$, since it is a proper subspace of $C[-1,1]$. If it were open, then there would exist an open ball $B(0,r) \subseteq \mathcal{P}[-1,1]$ around $0$. For any $f \in C[-1,1]$ we would have:
$$\frac{f}{\|f\|}\cdot \frac{r}{2} \in B(0,r) \subseteq \mathcal{P}[-1,1] \implies f = \frac{2\|f\|}{r} \cdot \left(\frac{f}{\|f\|}\cdot \frac{r}{2}\right) \in \mathcal{P}[-1,1]$$
since $\mathcal{P}[-1,1]$ is closed under scalar multiplication. This is a contradiction with the fact $\mathcal{P}[-1,1] \ne C[-1,1]$.
