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That is, we have to show that w.r.t. the metric $$d(p, q) = \sup_{0 \le t \le 1}|p(t) - q(t)|, $$ the polynomial is not complete.

I've been considering the sequence $$1, 1+x, 1 + x + \frac{x^2}{2!}$$ which I picked up as a hint to a problem similar to this, and trying to show this is a Cauchy sequence that converges to $e^x$. But I can't see how should I prove that the sequence is a Cauchy sequence and that it converges to $e^x$. Usually we have to index the terms of a sequence and then show that there exists some natural number $N$ such that $d(p_n, p_m) < \epsilon$ for all $m, n \ge N$, but I'm not sure how to do this with the sequence that I have. Please help.

Acid2
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    Do you know something about uniform convergence of a power series (strictly inside its radius of convergence)? – GEdgar Mar 31 '15 at 17:50
  • Note that you don't necessarily have to show that the given sequence converges to $e^x$, just that it doesn't converge to a polynomial, which might be simpler depending on what you know about Taylor series. – Santiago Canez Mar 31 '15 at 17:52
  • @GEdgar: We have discussed uniform convergence in class, yes. – Acid2 Mar 31 '15 at 17:57
  • Maybe you should think about $x^n$. – Karl Mar 31 '15 at 18:21

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The space of all polynomials is a normed space with the norm defined as \begin{equation} \|p\| = \sup_\limits{0\leq x\leq1}|p(x)|. \end{equation} which gives rise to the metric you are concerned with.

The following will show that the space of all polynomials on $[0,1]$ is not a Banach Space (Complete Normed Space) w.r.t any norm.

$\textbf{Lemma}:$ If $X$ is a Banach space then any Hamel Basis of $X$ is either finite or uncountable. $\textit{Proof}:$ If possible let $\{p_1,p_2,\ldots\}$ be a countable Hamel Basis for $X$. Define $A_n = Span\{p_1,\ldots,p_n\}$. Since each $A_n$ is a finite-dimensional subspace, it follows that $A_n$ is closed for all $n$. Further, $X = \cup_{n=1}^ {\infty}A_n$. Now by Baire Category Theorem, it follows that some $A_n$ must have non empty interior (since $A_n$'s are closed). But this is a contradiction as you can easily verify that any proper subspace of normed space has empty interior.

Since $\{1,x,x^2,\ldots\}$ is a countable Hamel basis for $P[0,1]$, it follows that $P[0,1]$ is not a Banach space w.r.t any norm.

This in particular shows that $P[0,1]$ is not a complete metric space w.r.t the metric you are considering.

Urban PENDU
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