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I want to say that $|\textbf{x}-\textbf y|<\delta$ implies $|x_1- y_1|<\delta$ and $|x_2- y_2|<\delta$ for a proof I am working on. This is assuming that $\textbf{x}=(x_1,x_2) \in \text R^2$ and $\textbf{y}=(y_1,y_2) \in \text R^2$. If true, I'd also like to extend this to $\textbf{x} \in \text R^{n_1+n_2}$ and $\textbf{x} \in \text R^{n_1+n_2}$ where $n_1 , n_2 \in \text N$.

I am wondering if this is obvious enough to state or would it be better to prove it? If I should prove it then what would be the best way? My guess would be to use the distance formula.

rioneye
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    I'm pretty sure it's obvious enough. Though since you're asking, you should probably write down the proof for yourself. – Karolis Juodelė Sep 22 '13 at 14:16
  • I guess I am asking if it is necessary to show something like this for a formal proof, or if you can assume that the reader knows that it is obvious enough. – rioneye Sep 22 '13 at 14:18

2 Answers2

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Note that $|x_1-y_1|^2\leq |x_1-y_1|^2+|x_2-y_2|^2$. The result follows from taking square roots on both sides.

azarel
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Recalling the fact

$$ \sqrt{a^2}=|a| \leq \sqrt{a^2+b^2}, $$

we have

$$ ||X-Y||=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} < \delta .$$

$$ \implies \sqrt{(x_1-y_1)^2} = |x_1-y_1| \leq ||X-Y|| < \delta.$$

Same for the other one.

Mhenni Benghorbal
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