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Take $y \in B_{\epsilon}(x)=\{r\in\mathbb{R}^n \mid ||x-r||<\epsilon\}$.

So I need to show that there exists $\bar{\epsilon}$ such that $B_{\bar{\epsilon}}(y) \subset B_{\epsilon}(x)$.

Take $z \in B_{\bar{\epsilon}}(y)$. So $||y-z||<\bar{\epsilon}$.

Then \begin{align*} ||x-z|| & = ||x-y+y-z|| \\ & \leq ||x-y|| + ||y-z|| \\ & < \epsilon + \bar{\epsilon} \end{align*}

but to be in $B_{\epsilon}(x)$ I need $||x-z||<\epsilon$ and this certainly isn't less than $\epsilon$.

Where is my mistake?

MHW
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1 Answers1

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Let $d$ be the distance between $x$ and $y\in B_{\epsilon}(x)$. Then let $\overline{\epsilon} = \frac{\epsilon -d }{2}$. Then for all $z\in B_{\overline{\epsilon}}(y),$ we have $||z - x || = ||z-y + y - x|| \leq ||z-y|| +||y - x|| \leq \frac{ \epsilon - d }{2} + d = \frac{ \epsilon + d }{2} < \epsilon$, since $d<\epsilon.$

user474330
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