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Here's the proof in my notes: enter image description here

Where does the last inequality come from? If I want to show that it's continuous at $((x,y)$ I can use the inverse triangle inequality to get $ (\|x^\prime\| + \|y\|)\varepsilon \leq (\|x\| + \|y \| + \varepsilon)\varepsilon$

Thanks.

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    I agree with h.h.543. The author even said \epsilon<1!2012-07-15

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Since $\varepsilon<1$, $\color{red}{\lVert x'\rVert}+\lVert y\rVert\leq \color{red}{\lVert x'-x\rVert+\lVert x\rVert} +\lVert y\rVert\leq\color{red}{\varepsilon}+\lVert x\rVert+\lVert y\rVert\leq \color{red}{1}+\lVert x\rVert+\lVert y\rVert.$

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    Thanks. May I ask a question which may be a bit off-topic? This proof shows that the multiplication is continuous with respect to the norm, without any reference to the norm condition $||xy||\le||x||||y||$ in the definition of normed algebra. So it means any algebra with a norm has continuous multiplication. Then my question is why do we need the norm condition in the definition of normed algebra ? I thought the norm condition is meant to make multiplication continuous.2017-11-19
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It comes from the triangle inequality $||x'||\leq ||x'-x||+||x||,$ and the hypothesis $||x'-x||\leq\varepsilon,\quad \varepsilon\in]0,1[$

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As Davide said,

$(\|x^\prime\| + \|y\|)\varepsilon \leq (\|x\| + \|y \| + \varepsilon)\varepsilon \leq (\|x\| + \|y \| + 1)\varepsilon$, as $\varepsilon < 1$, hence the given inequality.