Double Summation: Help needed with proof This question is about Theorem 8.3 of Baby Rudin:




what I don't understand is: 


*

*What is $E$? Is it subset of any metric space? Also, I can't see how $f_i$ continuous at $x_0$, because we are talking about only one sequence of set, converging to $x_0$.

*I have been struggling since yesterday to figure out why $$\lim\limits_{n\to\infty}\sum_{i=1}^{\infty}\sum_{j=1}^n a_{ij}=\lim\limits_{n\to\infty}\sum_{j=1}^n\sum_{i=1}^{\infty} a_{ij}$$ is permissible, and why $\lim\limits_{n\to\infty}\sum_{j=1}^n\sum_{i=1}^{\infty} a_{ij}$ indeed converges.
 A: 
(Convergence of $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty}a_{ij}$ (I))
Since $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{ij} = \sum_{i=1}^{\infty}f_i(x_0)$ by definition, and having supposed that $\sum_{j=1}^{\infty}|a_{ij}| = b_i \quad \forall i\in E$ (where E is our countable set), it is clear that $f_i(x_0) \leq  b_i$. Thus, $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty}a_{ij} \leq \sum_{i=1}^{\infty}b_i$. By assumption, $\sum b_i$ converges. Thus, (I) converges.
(Convergence of $\sum_{j=1}^{\infty} \sum_{i=1}^{\infty}a_{ij}$ (II)): Note that $\sum_{j=1}^{\infty} \sum_{i=1}^{\infty}a_{ij} = \lim_{n\to \infty}\sum_{j=1}^{n} \sum_{i=1}^{\infty}a_{ij} = \lim_{n \to \infty}\sum_{i=1}^{\infty} \sum_{j=1}^{n}a_{ij}$.
Since $\lim_{n \to \infty} \sum_{i=1}^{\infty} \sum_{j=1}^{n}a_{ij} \leq \sum_{i=1}^{\infty}b_i$, it is clear that (II) converges as well.

Since $\sum_{j=1}^{\infty}a_{ij} = f_i(x_0)$, it can be seen that $\sum_{j=1}^{\infty} \sum_{i=1}^{\infty}a_{ij} = \lim_{n \to \infty}\sum_{i=1}^{\infty} \sum_{j=1}^{n}a_{ij} = \lim_{n \to \infty}\sum_{i=1}^{\infty}f_i(x_n) = \lim_{n \to \infty} g(x_n) = g(x_0)$. By definition, $g(x_0) = \sum_{i=1}^{\infty}f_i(x_0)$. From (I), we know that $\sum_{i=1}^{\infty}f_i(x_0) = \sum_{i=1}^{\infty} \sum_{j=1}^{\infty}a_{ij}$.
Therefore, $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty}a_{ij} = \sum_{j=1}^{\infty} \sum_{i=1}^{\infty}a_{ij}$
A: $E$ is an abstract topological space. That is, as a set $E = \{x_0,x_1,x_2,...\}$, and the topology on $E$ is such that $x_n \to x_0$ in the topology of $E$. In other words, the open sets of the topology on $E$ ensure that $x_i, i>0$ are discrete points of $E$, and that $x_0$ is the only limit point.
A way of realizing $E$, is to look at it as a subset of another topological space, for example $E = \{0\} \cup \{\frac 1n : n \in \mathbb N\}$ as a subset of $\mathbb R$ qualifies. What are the open sets of $E$ as a topological space? By definition of the subspace topology, any open set of $E$ is an open set of $\mathbb R$ intersected with $E$. This gives you a description of a basis of open sets on $E$ : each $\{x_i\}$ is an open set in $E$ for $i > 0$, and each $\{0\} \cup \{\frac 1n : n \geq k\}$ is an open set around zero. 
Something similar occurs in the general case. More precisely, $E$ is  a first countable space, because every point has a countable local base. In  this case, it is known that continuity and sequential continuity are one and the same i.e. if we want to show that a function is continuous on $E$, then we don't need to look  at open sets : it will be enough to show that if $y_n \to y$ in $E$, then $f(y_n) \to f(y)$ . This will not be true if $E$ were not first countable. (Since $\mathbb R$ is first countable, this amounts to what is referred to as the "sequential definition of continuity" on the real line).
However, in $E$ there is only one non-trivial example of a convergent sequence (the rest are all constant sequences or a modification/subsequence of this sequence), namely $x_n \to x_0$. So, it will be enough to show $f(x_n) \to f(x_0)$ for continuity of $f$ on $E$.
$f_i,g$  are functions defined on $E$. To show that they are continuous on $E$, we only need to show the sequential definition of convergence, as I mentioned earlier.
So why is $f_i$ continuous? We need to check that $f_i(x_n) \to f_i(x_0)$, or that $\sum_{j=1}^n a_{ij} \to \sum_{j=1}^{\infty} a_{ij}$. However, the series $\sum_{j=1}^\infty a_{ij}$ is absolutely convergent, this is what $(12)$ is saying. Recall that this means that $\sum_{j=n+1}^\infty |a_{ij}| \to 0$ as $n \to \infty$. However, by the triangle inequality , $\left|\sum_{j=n+1}^\infty a_{ij}\right| \leq \sum_{j=n+1}^\infty |a_{ij}|$, so if the right hand side is going to zero as $n \to \infty$, so must the left side. Deduce from this that $f_i(x_n) \to f_i(x_0)$.
$g$ is a uniformly convergent sum of continuous functions, hence continuous (refer to the theorem given in Rudin's explanation).

In  the sequence of equalities at the end, it seems everything is following from the definitions of $f_i,g$ and their continuity, except for the following statement :
$$
\lim_{n \to \infty} \sum_{i=1}^\infty \sum_{j=1}^n a_{ij} =\color{red}{ \lim_{n \to \infty} \sum_{j=1}^n \sum_{i=1}^\infty a_{ij}}
$$
the contentious part being that the right hand side does not exist a priori. Indeed, what Rudin assumes trivial, the switching of an infinite  and a finite  summation, requires some work.
Suppose we show that for each $n$,  we have $\sum_{j=1}^n \sum_{i=1}^\infty a_{ij} = \sum_{i=1}^\infty \sum_{j=1}^n a_{ij}$, then the statement above would obviously follow. We know the right hand side is equal to $g(x_n)$. We have to show  the left hand side is equal to $g(x_n)$.
This follows from the fact that $$\sum_{i=1}^\infty a_{iJ} = \sum_{i=1}^\infty \left(\sum_{j=1}^J a_{ij} - \sum_{j=1}^{J-1} a_{ij}\right) = \sum_{i=1}^\infty \sum_{j=1}^J a_{ij} - \sum_{i=1}^\infty \sum_{j=1}^{J-1} a_{ij} = g(x_{J}) - g(x_{J-1})$$
(and equal to $g(x_1)$ if $J=1$). Therefore,  the left hand side is:
$$
\sum_{j=1}^n \sum_{i=1}^\infty a_{ij} = g(x_1) + \sum_{j=2}^n (g(x_j) - g(x_{j-1})) = g(x_n)
$$
Of course, since $\lim_{n \to \infty} g(x_n)$ exists and equals $g(x_0)$, the question about the limit existing is also answered. Finally, the proof is completed.
