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

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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  • $\begingroup$ 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. $\endgroup$ – Willie Wong Nov 10 '11 at 15:11
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    $\begingroup$ 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. $\endgroup$ – Willie Wong Nov 10 '11 at 15:12
  • $\begingroup$ 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" ? $\endgroup$ – Domates Nov 10 '11 at 15:55
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    $\begingroup$ 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. $\endgroup$ – Dylan Moreland Nov 10 '11 at 16:06
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    $\begingroup$ 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. $\endgroup$ – Arturo Magidin Nov 10 '11 at 17:06
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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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  • $\begingroup$ This has nothing to do with the Archimedean property. Just take $M$ to be $\|x-y\|/2$. $\endgroup$ – Michael Hardy Nov 10 '11 at 16:25
  • $\begingroup$ Ok. Thank you for answer. Well then , when $ M:=\min\|x-y\| $ , the condition that i want is not possible. for which $M$, $0<M\le\|x-y\| $ is true? $\endgroup$ – Domates Nov 10 '11 at 16:28
  • $\begingroup$ Shaun: Just edited your answer (replaced a faulty min by inf and rewrote the mathematical sentence in a more canonical way). Feel free to come back to the previous version if you like. $\endgroup$ – Did Nov 10 '11 at 16:54
  • $\begingroup$ @DidierPiau: Thanks! $\endgroup$ – Shaun Ault Nov 15 '11 at 15:51
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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.

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