I am asked to find the set of limit points of Z in R using the definition given below: Let (R,d) be a metric space. A point x belonging to R is called a limit point of Z if each open sphere centered on x contains atleast one point of Z other than x, in other words, if (B(x,r)-{x}) intersection with Z is null set. So if I choose any arbitrary x in R, then how should I show that the set of limit points is null set? Means What should I show using the definition? Does (B(x,r) -{x}) intersection with Z comes to be null set for every x in R?
Asked
Active
Viewed 603 times
1
-
Sorry for the mistake in definition. It is (B(x,r) -{x}) intersection with Z is not a null set. – Kavita Sep 26 '15 at 14:39
-
Repair your mistake by means of an edit. In general you should must detect wich for $x\in R$ some $r>0$ exists such that the intersection of $B(x,r)-{x}$ with $Z$ is empty. If you can find such $r$ then $x$ is not a limitpoint of $Z$. If you cannot then $x$ is a limitpoint of $Z$. – drhab Sep 26 '15 at 14:51
1 Answers
1
Yes, the set of all limit points of $\mathbb{Z}$ in $\mathbb{R}$ is empty; for, let $x \in \mathbb{R}$. If $x \in \mathbb{Z}$, then the deleted ball $B^{x}(\varepsilon) - \{ x \}$ of center $x$ and radius $\varepsilon$ has no intersection with $\mathbb{Z}$ for all $0 < \varepsilon < 1$; if $x \notin \mathbb{Z}$, let $d := \min \{ x - \lfloor x \rfloor, \lceil x \rceil - x \}$; then $B^{x}(\varepsilon ) - \{ x \}$ again has no intersection with $\mathbb{Z}$ for all $0 < \varepsilon < d$.
Yes
- 20,719
-
-
-
I am not getting what have you chosen d as . Is it greatest integer function [x] in d= min{ x-[x],[x]-x}? – Kavita Sep 26 '15 at 15:39
-
$\lfloor x \rfloor$ denotes the greatest integer lower bound of $x$; $\lceil x \rceil$ denotes the smallest integer upper bound of $x$. – Yes Sep 26 '15 at 15:47