Proposition:
$C\left ( \left [ a,b \right ],\mathbb{R} \right )$ is complete with respect to the metric induced by the uniform Norm $\left \| . \right \|_{0}$.
To prove this proposition, it is required that we show that every Cauchy sequence function in the metric space $C\left ( \left [ a,b \right ],\mathbb{R} \right )$ of continuous function converges.
However, in a proof from my notes, there is an addition demand and that is the requirement that every Cauchy sequence function in the set of continuous map $C\left ( \left [ a,b \right ],\mathbb{R} \right )$ is continuous.
Could someone kindly explain why this addition demand is required?
Thanks in advance.
To say that a metric $(M,d)$ is complete, that means that every sequence in $M$ that is Cauchy (according to $d$) is convergent (accroding to $d$) -- that is, has a limit $m \in M$. A priori, it does not even make sense for a sequence in a metric space to converge to something that does not lie in the metric space, because $d$ is not defined for it. Therefore, when you are talking about a single metric space such as $(C([a,b], \mathbb R),\|\cdot\|)$, and proving that every sequence in that metric space converges, in particular this convergence must be to a point in $C([a,b],\mathbb R)$.
In this particular case, it happens to be true that $(C([a,b]),\|\cdot\|)$ naturally lies inside another metric space, such as $(B([a,b]),\|\cdot\|)$, the bounded functions on $[a,b]$ with the sup-norm. Hence, the idea that a sequence inside $C([a,b])$ could converge to something outside $C([a,b])$ makes more sense. However, by default, convergence in a metric space is always to a point inside that metric space, unless specified otherwise.