In my notes, there is this theorem:
A sequence in $R^n$ converges (to a limit in $R^n$) iff it is Cauchy.
I understand that this theorem applies to all complete metric spaces, not just to $R^n$.
Now, here is my question:
The sequence of continuous functions $(f_n)$ in the space of continuous functions $C[0,1]$ equipped with the metric
$$d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$$
also converges to a limit in $C[0,1]$ iff it is Cauchy since $C[0,1]$ is complete.
But, what the convergence here explicitly is? Is it pointwise convergence or uniform convergence?
I understand that uniform convergence implies pointwise convergence with the same limit. Is it possible in the case above that there is a sequence of functions that converges pointwise and Cauchy in $C[0,1]$ but does not converge uniformly?
So is it more appropriate is it to re-write the Theorem as a sequence of continuous function in converges uniformly in $C$ iff it is Cauchy?
Thank you for the clarification.
Definitely it is meant uniform convegence since a uniform limit of a sequence of continuous functions is again continuous while a pointwise limit of continuous functions is not continuous in general. The convergence is always related to your topological structure of your space, and in your case, the topological structure is just a topology induced by metric. This metric is the supremum norm of $f-g.$
For example $f_n(x)=x^n$ on $[0,1]$ converges pointwise to $\chi_{\{1\}}$ which is not continuous.
I think that in some classical spaces you can "omit" the norm since the norm is by the definition known and other maybe still important norms does not play a great role in that given case. For example:
$l^p$ $(1\leq p<\infty)$ is a Banach spaces equipped by the norm $\|\cdot\|_p$. Since every sequence in $l^p$ is also bounded, you can also define the supremum norm on $l^p$. However, this norm does not imply completeness of the space $l^p$. Although it is worthwhile noticing that $l^p$ is dense in $c_0$ in the supremum norm.