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So, I'm working through Rudin's Principles of Mathematical Analysis, and I'm stuck on the proof of theorem 3.4, stated as follows.enter image description here

I'm working through the $(\Rightarrow)$ direction, that is, supposing that $\{\textbf{x}_n\}$ converges to $\textbf{x}$. I understand I can suppose I have an arbitrary $1\leq j \leq k$ and then suppose $\epsilon >0$. Since $\{\textbf{x}_n\}$ converges to $\textbf{x}$, there is $N\in \mathbb{N}$ such that for $n\geq N$, we have $|\textbf{x}_n - \textbf{x}|< \epsilon$. That is, $$\sqrt{(\alpha_{1,n}-\alpha_1)^2 + ... (\alpha_{k,n}-\alpha_k)^2} < \epsilon$$ I understand that this same $N$ should allow me, for $n\geq N$, to conclude that $|\alpha_{j,n}-\alpha_j|<\epsilon$ for the arbitrary $j$ by saying that $$|\alpha_{j,n}-\alpha_j|=\sqrt{(\alpha_{j,n}-\alpha_j)^2}\leq \sqrt{(\alpha_{1,n}-\alpha_1)^2 + ... (\alpha_{k,n}-\alpha_k)^2} < \epsilon$$ My trouble is in arguing that $$\sqrt{(\alpha_{j,n}-\alpha_j)^2}\leq \sqrt{(\alpha_{1,n}-\alpha_1)^2 + ... (\alpha_{k,n}-\alpha_k)^2}$$ completely rigorously. This is intuitive to me, but again, it's just the gritty details of a complete sound proof that really has me. It's probably easy, and I may just be having a brain fart. I was considering a simple inductive proof, but the whole inductive step would still depend on proving that for $x,c \in \mathbb{R}, c\geq0 $ we have that $|x|=\sqrt{x^2}\leq \sqrt{x^2+c}$. Can anyone help?

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    $\begingroup$ $x^2>0$ for all $x \in \mathbb{R}$ and $\sqrt{x}$ is a strictly increasing function (why?) $\endgroup$ – Matias Heikkilä May 9 '16 at 21:23
  • $\begingroup$ Oh, goodness of course! Thanks a bunch! $\endgroup$ – Sprinkle May 9 '16 at 23:29

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