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For any set $X \subset \mathbb{R}^d$, is it true that if $$||x|| < \sup\{||x_i|| : x_i \in X\} \quad\text{ and } \quad ||x|| > \inf\{||x_i|| : x_i \in X\}$$ then $$x \in X?$$

on-pasta
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    Do you mean $\sup{||x_i||}$? How do you define inequality on $\mathbb R^d?$ – Anvit May 28 '19 at 04:41
  • @Anvit Thanks for the catch. I accidentally wrote the element $x$ itself rather than the vector norm $||x||$, but now that's fixed. – on-pasta May 28 '19 at 04:42
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    @sawghol What does it mean to compare a scalar $||x||$ with a $d$-dimensional vector $\sup{x_i}$? – Darius May 28 '19 at 04:42

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You can just see for $d=1$ this is not true. Say your subset is $\{1,3\}$. Then $x=2$ satisfies hypothesis but conclusion doesn't follow

Anvit
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Not necessarily. Let d=1, and X={1/n | n /in N} then 0=inf(X)<2/3

Naman
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