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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?

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    Probably Kenneth R. Davidson, Allan P. Donsig: Real Analysis and Applications - Theory in Practice (Undergraduate Texts in Mathematics); [p.53](http://books.google.com/books?id=EPLJFwJv2pUC&pg=PA53).2012-11-13

2 Answers 2

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$\def\abs#1{\left|#1\right|}$(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 $\mathbb R^n$.

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    Still right. ${}{}{}$2012-11-13
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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 $k$.

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    Lol thank you very much for all of your insights2012-11-13