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If $0<\|x-y\|$ , can I say that there exists $0?

As a consequence of Archimedean property of $\mathbb{R} $, $ \exists M \in \mathbb{R} $ such that $ 0 < M < \|x−y\|$. Can I say that when any fixed $x\in \overline{A}=A\subset \mathbb{R}^{n}$ and any $y\in A^{c}$, if $ M:=\min\|x-y\|$, then $0 < M \le \|x−y\|$?

$\|\cdot\|$ : Euclidean norm on $\mathbb{R}^{n}$.

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    No: but for I suspect the "wrong" reason, because you don't know $\|x-y\| > 0$. Let $A$ be the open unit ball. Let $x = y$ be a point on the unit sphere. $x\in \bar{A}$ and $y\in A^c$, so $M = 0$, therefore $M >0$ is false.2011-11-10
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    Please edit your question to specify what $A$ is (or what properties the set $A$ possess, and also describe what $\min \| x - y\|$ means. As it is written, $x$ and $y$ are two give points, so I don't see why you are taking any minimum.2011-11-10
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    i have just edit question. since $x\in A $ and $y\in A^{c} $ ,then $x \neq y$. And from definition of norm $ 0 < \|x−y\| $. Is answer still "no" ?2011-11-10
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    It's still unclear to me what you're asking. If the minimum is being taken over all $x \in A$ and all $y \notin A^c$, then Prof Ault has answered your question (One should still be careful: the infinum of this expression is not attained by an actual pair $x, y$, so using "min" is dodgy). If you are fixing $y \in A^c$ and then minimizing over all $x \in A$ then there is something to be said.2011-11-10
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    If your set of possible pairs/values actually has a *minimum*, then you would be correct. The problem is that in general, $A^c$ is infinite, as is the set of values $\{\|x-y\|\mid y\in A^c\}$, in which case you do not know a priori that this set has a **minimum**. And while it *does* have an infimum, the strict inequality between $0$ and elements of the set does not guarantee the strict inequality between $0$ and the *infimum* of the set, just a non-strict one.2011-11-10

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