Rudin Theorem 3.4(a) 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?
 A: 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. 




A: Absolutely yes. In fact, if you read Rudin’s words, he’s not saying anything wrong. Fix some $\varepsilon$ then you will find such an $N$ such that (...). He skips the fact that this is actually the $N$ that allows all the coordinates to be smaller than the fixed epsilon.
About the first question, yes as well. In Analysis it is common to write the norm with the absolute value notation and in this case, if the context is clear, there is no confusion since the results of both the absolute values on the lhs and the norm of rhs are real numbers.
Note: Rudin’s textbook is pretty terse, and is probably not the best book where to find all the nitty gritty details of the proofs. But if you try yourself and then either come here and ask or check other references, then it’s one of the best books in introductory analysis.
