Problem with the limit of $x^n$ in $C[0,1]$ Let's consider a space of continuous function on the interval $[0, 1]$. We'll denote it by $C([0,1])$.
Let's define a sequence:
$$f_n(x) =\{x^n\}_{n=1}^{\infty}.$$
My task is to show that $f_n$ converges to a function $g: g \not\in C([0,1])$.
It's easy to calculate that:
$$g(x) = \begin{cases} 0,\text{if } x\in[0, 1)\\1, \text{if } x = 1\end{cases}.$$
To be honest I'm really confused. Because:


*

*I managed to prove that $C([0, 1])$ is complete,

*I know that a sequence of uniformly continuous functions converges to a continuous function, of course $x^n, \forall_{n \in \mathbb{N}}$ is a continuous function, moreover $[0,1]$ is a compact interval which means that $x^n$ is a uniformly continuous function.

 A: So the point is that we do indeed have $\lim_{n \to \infty} f(x) = g(x)$ for all $x \in [0, 1]$. However the usual topology on $C[0, 1]$ is induced by the supremum norm where
$$ \lVert f \rVert = \sup_{x \in [0, 1]} \lvert f(x) \rvert $$
Try showing that that $\lVert f_n - g \rVert = 1$ for all $n$ and the $g$ that you have defined, thus showing we do not have uniform convergence on $C[0, 1]$.
The two other statements you made are also true when we take the supremum norm on $C[0, 1]$. But $(f_n)$ is not a Cauchy sequence in $C[0, 1]$ so completeness is irrelevant and $(f_n)$ does not converge uniformly to $g$ so we won't expect $g$ to be continuous.
A: $f_n(x)$ does not converge in the sup norm, which is the norm on $C([0,1]):$ 
$f_n\to g$ pointwise, so if $f_n$ converges uniformly, then it must be to $g$.
For any fixed integer $n$, 
$\sup|f_n(x)- g(x)|=\sup|x^n-g(x)|>x^n$ for $x\neq 1$ and this will be $>1/2$ if $x$ is close enough to $1$. (in fact, the sup is equal to $1$ but we don't need that). 
We have thus found for each integer $n$, a number $0<x<1$ such that $\sup|f_n(x)-g(x)|>1/2$ and so $f$ does not converge to $g$ uniformly. 
