Is uniform convergence equivalent to Cauchy sequences for the space of continuous functions on an interval? Having received a perfect answer to this question, I am left trying to revisit what I thought I grasped at an intuitive level.
The present question asks for clarification on the comment:

The example is not using uniform convergence. A uniform limit of continuous functions is continuous. Because of that $C[a,b]$ is a Banach algebra when equipped with the infinity norm.

The broader concept for me to come to terms with is whether this example, which is clearly not a Cauchy sequence as per the answer to my prior question, is still a valid example for non-completeness, and if so, in what respect does it fail to contradict the idea that $C[a,b]$ is a Banach space.
This is the example (to avoid having to cross-reference the prior question):
$$f^{k}(x) = \frac 1 2 +\frac 1 \pi \tan^{-1}(kx)$$


The example is pretty bad. The lecturer wants to describe non-completeness and tries to use this example. From the little I saw in the video, he doesn't discuss what topology he is using. It looks like he is using the vector space of all continuous functions on the real line, which makes no sense as the "problem" with his example only  occurs at zero. And it looks like he is thinking of pointwise convergence.
With pointwise convergence, the same pathology can be found on $C[0,1]$ if one uses the usual example of $f_n(x)=x^n$.
