Proposition about Euclidean distance and open ball

Asked Sep 10 '20 at 16:10

Active Sep 10 '20 at 16:30

Viewed 33 times

About Euclidean space ($n$-dimensional)

$d_n(x,y)$ is Euclidean distance between $x$ and $y$.

For any $a\in\mathbb{R}^n$ and $r>0$ , $B(a,r) := \left\lbrace x\in\mathbb{R}^n : d_n(a,x)<r \right\rbrace$.

For any $b\in\mathbb{R}^n$ and $R>0$ , $B(b,R) := \left\lbrace x\in\mathbb{R}^n : d_n(b,x)<R \right\rbrace$.

Proposition ;

If $d_n(a,b)<R+r$ holds, then there exists c such that $c\in B(b,R) \cap B(a,r)$.

I drawn a picture and I found that such $c$ exists between $a$ and $b$. However, I couldn't represent such $c$ as a formula ($c=\ldots$).

How should I represent such $c$ as a formula?

edited Sep 10 '20 at 16:30

mwt

asked Sep 10 '20 at 16:10

daㅤ

Typeset mathematical terms using MathJax. Here's the tutorial. – SarGe Sep 10 '20 at 16:12
2

Consider the line segment between $a, b$ and the point on that line splitting the line segment in a ratio of $r:R$. – player3236 Sep 10 '20 at 16:12
@player3236 $c=a+(b-a) \frac{r}{dn(a,b)}$ $,,$ or $,,$ $c=b+(a-b) \frac{R}{dn(a,b)}$ is OK? – daㅤ Sep 10 '20 at 17:02
1

For the denominator, instead of $d_n(a,b)$, it should be $r+R$. – player3236 Sep 10 '20 at 17:10
However, when $c=a+(b-a) \frac{r}{r+R}$, $,,$ does $,,$ $dn(b,c)<R$ $,,$ hold? I can't prove that $,,$ $dn(b,c)<R$ $,,$ holds. – daㅤ Sep 10 '20 at 17:35

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