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?

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I think it would be nice if you mentioned the name of the book in your post. – Martin Sleziak Nov 13 '12 at 7:47
proof-theory is about an area of logic, see the [tag-wiki}math.stackexchange.com/tags/proof-theory/info), which explicitly says: This tag is NOT intended for questions asking how to write proofs or for checking a proof posted in the question. I have removed the tag. – Martin Sleziak Nov 13 '12 at 7:56
It was just an excerpt from notes I found online and I took a screenshot of it. – Hawk Nov 13 '12 at 8:12
Probably Kenneth R. Davidson, Allan P. Donsig: Real Analysis and Applications - Theory in Practice (Undergraduate Texts in Mathematics); p.53. – Martin Sleziak Nov 13 '12 at 9:36

2 Answers

$\def\abs#1{\left|#1\right|}$(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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(1) How did they come up with that inequality in the first place though. I know it holds, I am just not sure how they knew to use it like this. (2) So $k$ is what's changing here? – Hawk Nov 13 '12 at 7:50
(1) We have a bound on $\|x_k -a\|$ by assumption, we want to bound $|x_{k,i} - a_i|$ so it's a good idea to esitmate the latter by the former. (2) Yes. – martini Nov 13 '12 at 7:52
This is going to sound like a crazy question. If we replace the sequence with functions $f_k$ and we are mapping from $\mathbb{R^n}$ to $\mathbb{R^m}$, it would still be okay to choose $\dfrac{\epsilon}{m}$ because that is still fixed right? – Hawk Nov 13 '12 at 7:57
Sorry I meant to say 'm' and not 'n'. Because I am assuming I only have 'm' components of $f_k$ (so k < m) – Hawk Nov 13 '12 at 8:01
Still right. ${}{}{}$ – martini Nov 13 '12 at 8:03

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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(1) How do they know to even use that though? I know it holds, but I just don't see how I would know to use it like this (2) Ah okay gotcha, thank you – Hawk Nov 13 '12 at 7:51
This might be unusual if it is the first time you stumbled upon something similar, but I am sure that, with a little practise, you'll be doing things like this almost automatically very soon. "Young man, in mathematics you don't understand things. You just get used to them." (John von Neuman) – Martin Sleziak Nov 13 '12 at 7:59
Lol thank you very much for all of your insights – Hawk Nov 13 '12 at 8:11