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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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    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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Your first statement is missing some quantifiers, and should read (logicians out there, please forgive -- or correct -- any awkwardness in the following): $ \forall x \in \mathbb{R}^n,\ \forall y \in \mathbb{R}^n\setminus\{x\},\ \exists M>0,\ ||x - y||>M$ You see, the point is that the $M$ value depends on choices of $x$ and $y$. So when I look at various $x$, $y$ in some order, I may get a sequence of $M$'s that decrease to $0$. Indeed, if $A \subset \mathbb{R}^2$ is nonempty and $A \neq \mathbb{R}^2$, then $ \inf\{ ||x - y|| \;|\; x \in \overline{A}, y \in A^c\} = 0.$

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    @DidierPiau: Thanks!2011-11-15
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I think the sensible statement is that if $A \subset \mathbf{R}^n$ is closed and non-empty and $y \in A^c$, then \[ \min_{x \in A} |y - x| \] exists and is a positive number. Remember (or show!) that the function $\mathbf{R}^n \to [0, \infty)$ given by $z \mapsto |y - z|$ is continuous.