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As an exercise I am trying to show that we can find an injection from some open interval $(0,1)$ say into the open ball $B_{r}(x)$, where $r>0$ and $x \in \mathbb{R}^{2}.$

I'm a bit confused because I've only ever dealt with injective functions where the co-domain is $\mathbb{R}$ but here it is a ball.

But would the injection be something like each element in the open interval maps to a radius that is used in the definition of the open ball?

Thanks

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You can simply take$$\begin{array}{rccc}f\colon&(0,1)&\longrightarrow&B_r(x)\\&t&\mapsto&x+(tr,0).\end{array}$$It makes sense, since $\|f(x)-x)\|=\|(tr,0)\|=tr<r$. And it is injective, because$$(t_1r,0)=(t_2r,0)\iff t_1r=t_2r\iff t_1=t_2.$$

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  • $\begingroup$ is $t \in (0,1)$? The second line where you have $t$ arrow $x + (tr,0)$ is this meaning an element $t \in (0,1)$ maps to $x + (tr,0)$? Makes sense, and I was kind of on the right track! thanks for all the help recently! $\endgroup$ – Mathlearner Jun 19 '20 at 16:01
  • $\begingroup$ Yes, $t\in(0,1)$. I wrote that $f$ is a map from $(0,1)$ into $B_r(x)$. $\endgroup$ – José Carlos Santos Jun 19 '20 at 16:03
  • $\begingroup$ Yes I understand the $f$ is a map from $(0,1)$ into $B_{r}(x)$ but I'm not sure what the line underneath that means? $\endgroup$ – Mathlearner Jun 19 '20 at 16:05
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    $\begingroup$ The second line of the definition of $f$ means that $f$ maps $t$ into $x+(tr,0)$, yes. $\endgroup$ – José Carlos Santos Jun 19 '20 at 16:08

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