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For this proof, we're in $\mathbb{R}^n$ and $d$ is the Euclidean metric.

So the claim can be written $(|x-y|<r\Rightarrow |x'-y|<r')\Longleftrightarrow |x-x'|\leq r'-r$.

I think that I have ($\Leftarrow$):

Assume $|x-x'|\leq r'-r$.

Consider $y\in B(x,r)$.

Notice that $|x-x'|+|x-y|<|x-x'|+r\leq r'$.

Since $|(x'-x)+(x-y)|\leq |x'-x|+|x-y|< r'$, we have $|x'-y|<r'$.

Therefore, $y\in B(x',r')$ and $B(x,r)\subset B(x',r')$.

I've gone through 8 sheets of paper and as many hours on the other direction, but have yet to come up with anything substantial.

What I'm most confused about is how we can obtain the $-r$ term on the right side of the inequality.

HouseOfWinston
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1 Answers1

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For $\implies:$

By contradiction, assume $B(x,r)\subset B(x',r')$ and $|x-x'|>r'-r$. Now, let $y=x+\lambda\frac{(x'-x)}{|x'-x|}$, where $\lambda<r$. Then $|x-y|=\lambda<r$, so $y\in B(x,r)$. On the other hand, \begin{eqnarray} |x'-y| & = & \left|(x'-x)\left(\frac{\lambda}{|x'-x|}+1\right)\right|\\ & = & \lambda + |x'-x|\\ & = & \lambda + (|x'-x| - (r'-r)) + (r'-r)\\ & = & \lambda - r + (|x'-x| - (r'-r)) + r' \end{eqnarray}

Now, consider we take $$ \lambda = r-\frac{(|x'-x| - (r'-r))}{2} $$ (which is smaller than $r$), then we have \begin{eqnarray} |x'-y| & = & \lambda - r + (|x'-x| - (r'-r)) + r'\\ & = & \frac{(|x'-x| - (r'-r))}{2} +r'\\ & > & r' \end{eqnarray}

In short, if we assumed $|x-x'|>r'-r$, we have found some $\lambda\in (0,r)$ s.t. $y=x+\lambda\frac{(x'-x)}{|x'-x|}$ is in $B(x,r)$ but not in $B(x',r')$. This contradicts $B(x,r)\subset B(x',r')$.

Nate River
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