How to show that the space of polynomials is not complete Denote by $P[0,1]$ the set of all polynomials $p\colon [0,1]\to\mathbb{R}$; this is a vector space. Endow $P[0,1]$ with the norm $$\| p\|=\sup_{t\in [0,1]}{| p(t)|}.$$
I want to show that this particular norm on $P[0,1]$ isn't complete.
I think that, for example, the sequence $$p_n(x)=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^n}{n!}\qquad (n\in \mathbb{N})$$ is uniformly convergent on $[0,1]$ to the function $e^{x}$ that isn't a polynomial. Then $(p_n)$ is a Cauchy sequence, but it doesn't converge in $P[0,1]$. Now, how I can prove this, but more formally? 
Greetings... 

$\left\Vert p_{n}-p_{m}\right\Vert =\sup_{\left[  0,1\right]  }\left\vert
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{\displaystyle\sum\limits_{i=0}^{n}}
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\frac{x^{i}}{i!}-%
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{\displaystyle\sum\limits_{i=0}^{m}}
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\frac{x^{i}}{i!}\right\vert =\sup_{\left[  0,1\right]  }\left\vert
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{\displaystyle\sum\limits_{i=m+1}^{n}}
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\frac{x^{i}}{i!}\right\vert \leq%
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{\displaystyle\sum\limits_{i=m+1}^{n}}
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\frac{1}{i!}.$ Noting that $%
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{\displaystyle\sum\limits_{i=0}^{n}}
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\frac{1}{i!}\rightarrow e$, so $\left(  p_{n}\right)  $ is a Cauchy sequence.
Now, assume that $p_{n}\rightarrow p\in P\left[  0,1\right]  $. Hence
$p_{n}\left(  x\right)  \rightarrow p\left(  x\right)  $ $\forall x\in\left[
0,1\right]  .$ So  $p\left(  x\right)  =e^{x}\in P\left[  0,1\right]  .$
Assume $e^{x}=p\left(  x\right)  =%
%TCIMACRO{\dsum \limits_{i=0}^{M}}%
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{\displaystyle\sum\limits_{i=0}^{M}}
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a_{i}x^{i}$. Then $e^{x}=p^{\left(  M+1\right)  }\left(  x\right)  =0$ which
is impossible.
