Is $x^n$ Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$? Consider the sequence of functions
\begin{equation}
  f_n(x) = x^n, \quad x \in [0, 1].
\end{equation}
Is this sequence Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$?
The pointwise limit is not continuous because we have
\begin{equation}
  f(x) := \lim_{n \to \infty} f_n(x) =
  \begin{cases}
    0 & \mbox{if $x \in [0, 1)$} \\
    1 & \mbox{if $x = 1$}
  \end{cases}
\end{equation}
Moreover we have
\begin{align}
  \|x^n - x^m\|_{\infty} &= \sup_{x \in [0, 1]} |x^n - x^m| \\
  &\leq \sup_{x \in (0, 1)} |x^n| + |x^m| \\
  &\leq 2 \varepsilon.
\end{align}
The first inequality uses the triangle inequality and the fact that $x^n - x^m = 0$ for $x \in \{0, 1\}$. The second inequality uses the fact that for $x^n \to 0$ for $x \in (0, 1)$ so we can find an $N$ such that $x^n < \varepsilon$ for any $n > N$.
Thus $f_n$ is Cauchy in $(C[0, 1], \|\cdot\|_{\infty})$, which is complete. Hence $f_n \to f \in (C[0, 1], \|\cdot\|_{\infty})$. But we already showed that $f$ is not continuous. Contradiction.
Can anyone tell me where things are going wrong?
 A: We have $\sup_{x\in (0,1)}|x|^n=1$, so the argument in the OP doesn't work. An other reason it that the sequence $\{f_n\}$ is not Cauchy for the supremum norm. Indeed, if it was the case, then $\sup_{x\in (0,1)}|x^{2n}-x^n|$ would be $0$. Let $g_n(x):=x^n-x^{2n}=x^n(1-x^n)$. If $x^n=1/2$, $x=2^{-1/n}$  so $\sup_{x\in [0,1]}|x^{2n}-x^n|\geq \frac 14$ (it's actually equal but we don't need that). 
A: The inequality
$$|x^n|+|x^m|\le 2\varepsilon$$
is true for any fixed $x\in(0,1)$ and for any given $\varepsilon>0$, starting from some $N$ (i.e. for all $n,m\ge N$).
But the number $N$ depends on $x$. (You are using the fact that $x^n\to 0$ for $x\in (0,1)$, but this is only pointwise convergence, not uniform.)
So you don't get 
$$\sup_{x\in(0,1)}|x^n|+|x^m|\le 2\varepsilon$$
for $m,n>N$ from the above inequality. (In order to get this you would need to have the same $N$ for each $x\in(0,1)$.)
A: No, the sequence is not Cauchy $(C[0, 1], ||\cdot||_{\infty})$. A formal proof is given below.
Proof. By definition if $x^n$ is Cauchy, then for all $\varepsilon > 0$ there exists $N \in \mathbb{N}$ such that
\begin{equation}
  ||x^n - x^m||_{\infty} < \varepsilon
\end{equation}
whenever $n, m \geq N$.
Fix $\varepsilon > 0$. Suppose for a contradiction that such an $N$ exists. Then in particular we can set $n = N$ and it must be the case that for all $m \geq N$
\begin{equation}
  ||x^N - x^m||_{\infty} < \varepsilon.
\end{equation}
But as $x^N$ is continuous and equal to $1$ when $x = 1$. Thus for any $\eta > 0$ we can find a $\delta > 0$ such that $|1 - x^N| < \eta/2$ whenever $|1 - x| < \delta$. But this means $x^N > 1 - \eta/2$. Let $x_0$ be arbitrary in $(1-\delta, 1)$. At $x_0$ there exists $m$ such that $x_0^m < \eta/2$. Thus at $x_0$ we have
\begin{equation}
  |x_0^N - x_0^m| = x_0^N - x_0^m \geq 1 - \eta/2 - \eta/2 = 1 - \eta.
\end{equation}
Choose $\eta$ so that $1 - \eta > \varepsilon$. Then
\begin{equation}
  ||x^n - x^m||_{\infty} = \sup_{x \in [0, 1]} |x^n - x^m| \geq |x_0^n - x_0^m| \geq \varepsilon.
\end{equation}
A contradiction. Thus $x^n$ is not Cauchy in $(C[0, 1], ||\cdot||_{\infty})$.
A: Note that your space is complete, so a Cauchy sequence must converge. But the limit of your sequence is not continuous, hence it cannot lie in your space. Hence the sequence is not Cauchy.
A: The Cauchy sequence in C[0,1] is not point wise convergent to a continuous function in C[0,1]. Thus, C[0,1] is NOT complete under point-wise convergence. 
