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 &amp; \mbox{if $x \in [0, 1)$} \\
    1 &amp; \mbox{if $x = 1$}
  \end{cases}
\end{equation}

Moreover we have

\begin{align}
  \|x^n - x^m\|_{\infty} &amp;= \sup_{x \in [0, 1]} |x^n - x^m| \\
  &amp;\leq \sup_{x \in (0, 1)} |x^n| + |x^m| \\
  &amp;\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 $x^n &lt; \varepsilon$ for any $n &gt; N$.

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?
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).
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.