How to prove or disprove the following set is compact?

Question

$P$ is set of all real polynomial in one variable. Define $$d(p,q)=\max\{|p(x)-q(x)|:x \in[0,1]\}$$ and $$K=\{p \in P,d(p,0)\leq 1 \}$$Prove or disprove $K$ is compact.

I think the only way to prove this is use the definition (open cover has finite subcover), but I don't know how.

Answer 1

No, $K$ is not compact, because $P$ (and hence $K$) is not complete under the metric $d$ by e.g., Stone-Weierstrass. Note that a metric space is compact if and only if it is complete and totally bounded.

Answer 2

You can show that $K$ is not compact by finding a sequence of polynomials $(p_n)_n$ in this set such that $d(p_n,p_m) \ge \delta > 0$ for all $n,m$. The sequence of Chebyshev polynomials might do the trick.

Answer 3

Let $p_n(x)=e^{-1}\sum_{j=0}^n x^j/j!$ for $x\in [0,1].$ Then $\{p_n:n\in \Bbb N\}\subset K,$ and $(p_n)_{n\in \Bbb N}$ converges uniformly to $e^{x-1}$ on $[0,1].$

But $f(x)=e^{x-1}$ is not a polynomial. Because $f'=f,$ but the derivative of a non-constant polynomial of degree $D$ is a polynomial of degree $D-1.$

So $K$ is not sequentially compact, hence not compact.