If $x,z$ $\epsilon$ $R$, show that for every $\epsilon >0$ there is a $\delta > 0$ such that if $y$ $\epsilon$ $R$ satisfies $|y-x|< \delta$ then $|zy - xz| < \epsilon$.
So I tried to take $\epsilon =1$ and I tried to solve from there, but I couldn't seem to make it work. Any ideas/help?
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Zoë Soriano
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Be careful: it says for every $\epsilon>0$. Finding a $\delta$ for a single $\epsilon$ (e.g., $\epsilon = 1$) is not sufficient. – Dec 15 '14 at 03:35
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if $z = 0$, the statement is a triviality. Hence, we assume $z \neq 0 $.
Let $\epsilon >0 $ be given. Take $\delta = \frac{\epsilon}{|z|} $. Let $y \in \mathbb{R}$ be arbitrary and let us suppose that $|y -x | < \delta$. It follows that
$$ |zy - zx| = |z(y-x)| = |z||y-x| < |z| \cdot \frac{ \epsilon}{ |z|} = \epsilon$$