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Following a reference from “Elementos de Topología general” by Angel Tamariz and Fidel Casarrubias.

Let be $a\in\mathbb{R}\setminus\{0\}$ and we choose $\epsilon>0$. We define $\delta=\min\left\{\frac{|a|}2,\frac{\epsilon|a|^2}2\right\}$ and so we observe that if $y\in B(a,\delta)$, then $$ \left|\frac{1}y-\frac{1}a\right|=\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|<\frac{2}{|a|^2}\frac{|a|^2\epsilon}2=\epsilon $$

and so we conclude that $\frac{1}x$ is continuous.

Unfortunately I don't understand why $\frac{|y-a|}{|y||a|}\le\frac{2}{|a|^2}|y-a|$, that is $\frac{1}{|y|}\le\frac{2}{|a|}$. In fact we have $$ |y-a|<\delta\le\frac{|a|}2\Rightarrow-2|a|<a-\frac{|a|}2<y<a+\frac{|a|}2<2|a|\Rightarrow|y|<2|a|\Rightarrow\frac{1}{|a|}<\frac{2}y $$ that is in contradiction with $\frac{1}{|y|}\le\frac{2}{|a|}$. Could someone help me, please?

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Since $\lvert y-a\rvert<\frac{\lvert a\rvert}2$, we have$$\lvert y\rvert\geqslant\bigl\lvert\lvert a\rvert-\lvert y-a\rvert\bigr\rvert=\lvert a\rvert-\lvert y-a\rvert>\frac{\lvert a\rvert}2.$$So$$\frac{\lvert y-a\rvert}{\lvert y\rvert\lvert a\rvert}<\frac{\lvert y-a\rvert}{\lvert a\rvert}\times\frac2{\lvert a\rvert}=\frac2{\lvert a\rvert^2}\lvert y-a\rvert.$$

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  • $\begingroup$ Sorry, could you prove that $\lvert y\rvert\geqslant\bigl\lvert\lvert a\rvert-\lvert y-a\rvert\bigr\rvert=\lvert a\rvert-\lvert y-a\rvert>\frac{\lvert a\rvert}2$? Unfortunately it seems that it is difficult to me. $\endgroup$ Mar 4 '20 at 23:21
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    $\begingroup$ Note that$$\lvert a\rvert=\lvert a-y+y\rvert\leqslant\lvert a-y\rvert+\lvert y\rvert$$and that therefore $\lvert y\rvert\geqslant\lvert a\rvert-\lvert y-a\rvert$. Now, use the fact that $\lvert y-a\rvert<\frac{\lvert a\rvert}2$. $\endgroup$ Mar 4 '20 at 23:24
  • $\begingroup$ Sorry, but I didn't succeed: I only see that $|y-a|<\frac{|a|}2<|a|\Rightarrow 0<|a|-|y-a|$. Could you explain, please? $\endgroup$ Mar 4 '20 at 23:35
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    $\begingroup$ Since $\lvert y-a\rvert<\frac{\lvert a\rvert}2$, then $\lvert a\rvert-\lvert y-a\rvert>\lvert a\rvert-\frac{\lvert a\rvert}2=\frac{\lvert a\rvert}2.$ $\endgroup$ Mar 4 '20 at 23:38
  • $\begingroup$ Okay, now it is clear. Thanks! $\endgroup$ Mar 4 '20 at 23:42
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$$ |a| = |y|\leq |a-y| < \delta < \frac{|a|}{2}$$

Thus

$$ |y| > \frac{|a|}{2}$$

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