I would like to ask whether the following proof is correct and, in particular, why the step (*) of this proof is valid?

The problem is to show that the function $+: \mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto x+y$ is uniformly continuous and give a counterexample for $\times: \mathbb{R}^2 \to \mathbb{R}, (x,y)\mapsto xy$.

addition $f(x,y)=x+y$

Given $\varepsilon>0,$ let $\delta=\varepsilon/2.$ Then for any $(x,y)$, $(u,v)$,


$\implies|x-u|<\varepsilon/2\,,\,|y-v|<\varepsilon/2$ (*)

$\hspace{0ex}\implies d(f(x,y),f(u,v))=|x+y-(u+v)|$



multiplication $g(x,y)=xy$

Given $(x,y)\,, x>0,y>0,$ and $\delta>0,$ observe that




Consequently, given $\varepsilon>0\,,\,\,$ for any $\delta>0$ , if we choose x and y positive with $x+y>\varepsilon/\delta\,\,,\, $then $|xy-(x+\delta)(y+\delta)|>\varepsilon.$

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    $\begingroup$ The argument is OK, but at the end you had to say " we choose $x$ and $y$ positive with $x+y >\epsilon / \delta$ and $|x-y| <\delta$" (say with $x=n, y=n+\delta /2$ with $n$ large enough). $\endgroup$ – Kavi Rama Murthy Sep 4 '18 at 8:45
  • $\begingroup$ But how does $d((x,y),(u,v))<\varepsilon/2$ imply $|x-u|<\varepsilon/2\,,\,|y-v|<\varepsilon/2$ in the second step? This does not follow using the Euclidean metric.. $\endgroup$ – Cebiş Mellim Sep 4 '18 at 8:48
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    $\begingroup$ $|x-u| \leq \sqrt {(x-u)^{2}+(y-v)^{2}} <\delta$. Similarly $|y-v| <\delta$. $\endgroup$ – Kavi Rama Murthy Sep 4 '18 at 8:49

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