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Can someone explain Lemma 2.16 to me?

What is $\mathbf{R}_\delta$? And how does step (2.18) prove the whole thing is one to one?

Guess: I am guessing (2.18)'s conclusion is somehow related to Mean Value Thrm? A wild guess with no basis

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up vote 2 down vote accepted

$R_{\delta}$ is defined in the first few lines of the pdf file. Establishing 2.81 shows $f$ is injective in $R_{\delta}$ because if $f(x)=f(y)$ then $|x-y|\leq 2 |f(x)-f(y)|=0$ so $x=y.$

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Oh so if $f(x) = f(y)$. But if that were the case, why not just use the Mean Value Theorem to prove this in the first case? The formulas look nearly identical – love Feb 20 '13 at 23:59
@love I am unsure of what you are referring to exactly. Can you cite an equation number? – Ragib Zaman Feb 21 '13 at 0:02
A function is one to one if $f(x)=f(y) \implies x=y.$ In 2.81, if we suppose $f(x)=f(y)$ we can indeed conclude $x=y$ by putting our information in. But if you try that in the one you just mentioned, you only get $0 \leq |f'(p)||x-y|$ which doesn't let you conclude $x=y.$ – Ragib Zaman Feb 21 '13 at 0:10
Whether $|f'(p)|$ is zero or not, having $0\leq |f'(p)||x-y|$ doesn't let us say $x=y.$ We might have $|x-y|=1$ and the inequality will still be true. In fact, $0\leq |f'(p)| |x-y|$ is trivially true since the right hand side is non-negative because of the absolute values. I don't understand how the MVT will help here. – Ragib Zaman Feb 21 '13 at 0:18
Could you tell me how 2.82 shows it is onto? Why does 1 > 1/2 imply this? – love Feb 21 '13 at 0:43

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