How to show something is a Banach space Show that 
$$(C^0([0,1]),\|\cdot\|_\infty) \quad\text{where} \|f\|_{\infty} :=\max_{[0,1]}\|f(x)\|$$
 is a Banach space
I have tried to go about showing that $\|\cdot\|_{\infty}$ is a norm using the axioms of a norm, but now I cannot figure out how to show that the space is a complete normed space.
 A: Suppose $(f_n)$ is Cauchy in $C^0([0,1])$. Then for each $x\in[0,1]$ we have
$$
|f_n(x)-f_m(x)| \leq \|f_n-f_m\|_\infty.
$$
Thus $(f_n(x))$ is Cauchy in $\mathbb{R}$ for each $x\in[0,1]$. Since $\mathbb{R}$ is complete, there exists $f:[0,1]\to\mathbb{R}$ such that $f(x) = \lim_n f_n(x)$ for each $x\in[0,1]$. It remains to show that $f\in C^0([0,1])$ and $f_n\to f$ in $C^0([0,1])$.
To see that $f$ is continuous, fix $x_0\in[0,1]$ and let $\varepsilon>0$. By the fact that $(f_n)$ is Cauchy, we can choose $N\in\mathbb{N}$ such that $\|f_n-f_m\|_\infty < \varepsilon/3$ for all $n,m\geq N$. Since $f_N$ is continuous at $x_0$, there exists $\delta>0$ such that $|x-x_0|<\delta$ implies
$|f_N(x)-f_N(x_0)|<\varepsilon/3$. Thus if $|x-x_0|<\delta$ we have
\begin{align*}
|f(x)-f(x_0)|
&=\lim_n|f_n(x)-f_n(x_0)| \\
&\leq \lim_n|f_n(x)-f_N(x)|+\lim_n|f_N(x)-f_N(x_0)|+\lim_n|f_N(x_0)-f_n(x_0)| \\
&< \varepsilon.
\end{align*}
Now we just need to show that $f_n\to f$ in $C^0([0,1])$.
Given $\varepsilon>0$, choose $N\in\mathbb{N}$ such that $\|f_n-f_m\|_\infty<\varepsilon$ for all $n,m\geq N$. For every $x\in[0,1]$ and $n\geq N$ we have
$$
|f_n(x)-f(x)| = \lim_m|f_n(x)-f_m(x)|\leq \limsup_m\|f_n-f_m\|_\infty \leq \varepsilon.
$$
Therefore $\|f_n-f\|_\infty\leq \varepsilon$ for every $n\geq N$.
