The set of Polynomials is not complete in sup-norm over [0,1] Let $P$ be the set of all polynomials and consider the sup-norm on $[0,1]$ by $$||p||_\infty = \sup\{|p(x)| : x \in [0,1]\}.$$ I need to show $P$ is not complete in the sup-norm on the interval $[0,1]$. To do this, it suffices to show there exists a sequence of elements in $P$ which is Cauchy but has no limit in $P$. To this end, consider the sequence of polynomials $(p_n)$ defined by $$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k},$$ where $x \in [0,1]$. I have already proved that $(p_n)$ is in fact Cauchy with respect to the sup-norm, but I am confused on how to show that $$p(x) = \frac{1}{1 - \frac{x}{2}}$$ is the only possible limit for the sequence $(p_n)$. We know that $$\lim_{n \to \infty} p_n(x) = \lim_{n \to \infty} \sum_{k=0}^{n} \frac{x^k}{2^k} = \sum_{k=0}^{\infty} \left(\frac{x}{2}\right)^k = \frac{1}{1 - \frac{x}{2}} = p(x)$$ for $x \in [0,1]$. Also, $\sum_{k=0}^\infty \left(\frac{x}{2}\right)^k = \sum_{k=0}^\infty \frac{1}{2^k} x^k$, is a power series with radius of convergence $(-2,2).$ Therefore, since $[0,1] \subseteq (-2,2)$, it follows that $\sum_{k=0}^\infty \left(\frac{x}{2}\right)^k = \sum_{k=0}^\infty \frac{1}{2^k} x^k$ converges uniformly on $[0,1]$ and is continuous on $[0,1]$. I believe this would then imply that $p(x)$ is in the larger space $C([0,1])$ and hence, $p(x)$ is the only possible limit for our Cauchy sequence $(p_n)$. Is this argument valid? Do I even need to state any of this? Thanks.      

Indeed you are trying to show that set of all polynomials complete! you have to show that the limit of a series of polynomial is not a polynomial. One of the ways is proving by counterexample. We have
$$\lim_{n\rightarrow \infty} 1-x^2/2+x^4/4-\cdots=cosx$$  which is not polynomial then the space is not complete.
