Question on a proof of a sequence 
I have some questions
1) In the forward direction of the proof, it employs the inequality $|x_{k,i} - a_i| \leq (\sum_{j=1}^{n} |x_{k,j} - a_j|^2)^{\frac{1}{2}}$. What exactly is this inequality?
2) In the backwards direction they claim to use the inequality $\epsilon/n$. I thought that when we choose $\epsilon$ in our proofs, it shouldn't depend on $n$ because $n$ is always changing?
 A: $\def\abs#1{\left|#1\right|}$(1) We have that 
$$ \abs{x_{k,i} - a_i}^2 \le \sum_{j=1}^n \abs{x_{k,j} - a_j}^2 $$
for sure as adding positive numbers makes the expression bigger. Now, exploiting the monotonicity of $\sqrt{\cdot}$, we have
$$ \abs{x_{k,i} - a_i} = \left(\abs{x_{k,i} - a_i}^2\right)^{1/2} \le \left(\sum_{j=1}^n \abs{x_{k,j} - a_j}^2\right)^{1/2} $$
(2) When you talk about sequences $(x_n)$, where you use $n$ to index the sequence's elements, your $\epsilon > 0$. But in this case, $n$ denotes the (fixed, not chaning for different elements $x_k$) dimension of $\mathbb R^n$, are you are talking about a sequence $(x_k)$ in $\mathbb R^n$.
A: 
1) In the forward direction of the proof, it employs the inequality $|x_{k,i} - a_i| \leq (\sum_{j=1}^{n} |x_{k,j} - a_j|^2)^{\frac{1}{2}}$. What exactly is this inequality?

Obviously, for any $i$ you have
$$|x_{k,i}-a_i|^2 \le \sum_{j=1}^{n} |x_{k,j} - a_j|^2.$$
(You are working with a sum of non-negative numbers, one summand cannot be larger than the whole sum.)
Therefore
$$|x_{k,i}-a_i|=\sqrt{|x_{k,i}-a_i|^2} \le \sqrt{\sum_{j=1}^{n} |x_{k,j} - a_j|^2}.$$

2) In the backwards direction they claim to use the inequality $\epsilon/n$. I thought that when we choose $\epsilon$ in our proofs, it shouldn't depend on $n$ because $n$ is always changing?

In the whole proof $n$ is fixed - it is the dimension of $\mathbb R^n$; the variable used for indices in the sequence is $k$.
