I am reading Henri Cartan's Elementary Theory of Analytic Functions of One or Several Complex Variables. On p.18, he wrote:
Recall that the limit of a uniformly convergent sequence of continuous functions (on a topological space $E$) is continuous. In particular, the sum of a normally convergent series of continuous functions is continuous. An important consequence of this is:
PROPOSITION 1.2. Suppose that, for each $n$, $\lim_{x\to x_0}u_n(x)$ exists and takes the value $a_n$. Then, if the series $\sum_nu_n$ is normally convergent, the series $\sum_na_n$ is convergent and $$\sum_na_n = \lim_{x\to x_0}\left(\sum_nu_n(x)\right)$$ (changing the order of the summation and the limiting process).
I understand why the proposition is true (a proof under the weaker assumption that $\sum_nu_n$ converges uniformly can be found in theorem 7.11 of "Baby Rudin", and normal convergence implies uniform convergence). What I don't understand, however, is Cartan's remark.
Why is this proposition a consequence of the fact that the sum of a normally convergent series of continuous functions is continuous? The $u_n$s in the proposition are not necessarily continuous. Am I missing something?
For simplicity, let's assume $u_n : \mathbb{R} \to \mathbb{R},$ and since the proposition doesn't involve $u_n(x_0)$ at all, we may define, without loss of generality, $u_n(x_0) = a_n$ in order to assume the $u_n$ are continuous at $x_0.$
Since Henri Cartan's remark holds for a general topological space $E,$ we may define $E = (\mathbb{R}, \tau)$ with the topology $$\tau = \{A \cup B(x_0,r) : A \subseteq \mathbb{R}\setminus\{x_0\}, r \geq 0\} \\ \text{where}\quad B(x_0,r) = \{x \in \mathbb{R} : |x - x_0| < r\}$$ and then note that $u_n : E \to \mathbb{R}$ is continuous if and only if $u_n$ is continuous at $x_0$ in the usual topology.
With this choice of $E,$ the remark now implies the proposition.