I believe I understand the proof, but I just want to be sure I understand fully what Rudin is saying.
The theorem is:
Suppose $x_n \in \mathbb{R}^k$ and $x_n = (a_{1,n}, \ldots, a_{k,n})$ Then $\{x_n\}$ converges to $x = (a_1, \ldots, a_k)$ if and only if $\lim\limits_{n \to \infty}$ if and only if $\lim\limits_{n \to \infty} a_{j,n} = a_j$ for $1 \leq j \leq k$.
The forward direction is more or less clear, though I have only one small question.. Rudin writes that
$$
|a_{j,n} – a_j| \leq |x_n -x|.
$$
I assume the right side is a vector norm, the right-hand side is the absoltue value of a difference of scalars, because the inequality comes from taking the square root of a square. Is that right?
The backward direction is a bit more confusing. Rudin's proof, replicated verbatim, is:
Conversely, if (2) holds, then to each $\epsilon > 0$ there corresponds an integer $N$ such that $n \geq N$ implies $|a_{j,n} – a_j| < \frac{\epsilon}{\sqrt{k}}$. Hence $n \geq N$ implies
$$
|x_n – x| = \left(\sum\limits_{j=1}^k |a_{j,n} – a_j|^2 \right)^{1/2} < \epsilon,
$$
so that $x_n \to x$.
Here is my confusion: I think Rudin has skipped a step. He picks an $N$ for only a single $j$, but not for each $j$. It seems to me that we should, for each $j$, pick $N_j$ so that $n \geq N_j$ implies $|a_{j,n} – a_j| < \frac{\epsilon}{\sqrt{k}}$ and then set $N = \max(N_1, \ldots, N_k)$. Otherwise, Rudin is in some way bounding the above sum by a single index, which seems peculiar.
Would I be correct that Rudin has actually done what I just suggested, but been silent about it in the write-up?
Best Answer
You are absolutely thinking right for converse. When i was also dealing with this question , i also quitioned the same with my proffessor.
And for the forward direction , if $|x_n - x |$ < e , then you can expand xn and x in k touples . Then take it's absolute value , Square both side. Then a inequality comes . And after that each element of touple will also follow the same inequality and you will able to proof the theorem.
I am attaching some of pictures here , here i am give the proof of this in detail.