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Suppose we work in an euclidean space. Let $B(x,r)$, $B(y,s)$ two balls with radius r and s, respectively. If $B(x,r)$ is a subset of $B(y,s)$ then $d(x,y)\leqslant s-r$.

I tried it and I saw that intuitively it is true that if $d(x,y) > s-r$ then $B(x,r)$ is not a subset of $B(y,s)$ because there is at least one element $w'$ such that $d(w',y)\geqslant s$.

I would appreciate any help to express this thought more formally and to reach a contradiction.

BigbearZzz
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Marios Gretsas
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1 Answers1

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Let $n$ be a strictly positive integer. Write $z=x+(r-1/n){{x-y}\over{\|x-y\|}}$, $d(x,z)=\|x-z\|=\|(r-1/n){{x-y}\over{\|x-y\|}}\|=r-1/n$. We deduce that $z\in B(x,r)\subset B(y,s)$. $d(y,z)=\|x+(r-1/n){{x-y}\over{\|x-y\|}}-y\|=(1+{{r-1/n}\over{\|x-y\|}})\|x-y\|=\|x-y\|+r-1/n\leq s$.

This implies that $\|x-y\|\leq s-r+1/n$ for every $n\in N-\{0\}$ thus $\|x-y\|\leq s-r$.

Tsemo Aristide
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