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Let $(X,d)$ be a metric space ,$\Omega$ be a open subset of $X$ and $x_1,x_2 \in \Omega$ with $x_1 \neq x_2$ and put $d(x_i)=d(x_1, X \setminus \Omega), i=1,2$ suppose that $$2^{k-1} < d(x_i) \leq 2^{k}$$ for a given $k \in \mathbb{Z}$, and exists $y \in B(x_1, \frac{1}{4} d(x_1)) \cap B(x_2, \frac{1}{4} d(x_2)))$ then show that $$d(x_1,x_2) \geq \frac{1}{20} \min \{d(x_1),d(x_2)\}$$ I try to make the proof by contradiction supposing that $d(x_1,x_2) < \frac{1}{20} \min\{d(x_1),d(x_2)\}$ and using triangle inequality and the relations given but i not sure to these is correct, any hint or help i will be very grateful

Nick Weber
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    I don't think that you have included all things here... If $x_1=x_2$, then all the conditions should be met but the conclusion is clearly wrong... Maybe I'm missing something. But, could you check again? – Feng Aug 24 '23 at 00:53
  • @Feng sure i forget to put the condition that $x_1 \neq x_j$ – Nick Weber Aug 24 '23 at 00:55
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    This is still strange. Let's say $X=\mathbb R$ and $\Omega=(0,2)$. Given an arbitrary $\epsilon\in(0,0.1)$, consider $x_1=1, x_2=1+\epsilon$, then $d(x_1)=1, d(x_2)=1-\epsilon$ so we can take $k=0$. It is easy to verify the remaining condition. However, the conclusion doesn't hold because $d(x_1, x_2)=\epsilon$ can be very very small. – Feng Aug 24 '23 at 01:03
  • @Feng in fact i thing to the inequality is estrange, i reading one proof about the Whitney descomposition theorem in doubling metric spaces and that inequality appears, i try to proving but i cant – Nick Weber Aug 24 '23 at 01:06

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