Question #308343

Give an example for each of the following:


A set having no limit point.

1
Expert's answer
2022-03-13T18:07:59-0400

Definition. A limit point (accumulation point) of a set SS in a metric space XX is a point xXx\in X any puncted neighborhood of which contains an element of SS. The set of all limit points of SS is denoted by SS'.


In other words, SS' is the set of all points xXx\in X at which the set SS is locally infinite. Really, if there exists a neighborhood UU of xx such that USU\cap S is finite then, by putting r=min{dist(x,y):yUS,yx}r=\min\{dist(x,y): y\in U\cap S,y\ne x\}, we have that a puncted open ball B˙r(x)\dot{B}_r(x) contains no elements of SS. This means that xx is not a limit point of SS. Conversely, if any neighborhood UU of xx contains an infinite subset of SS then a puncted neighborhood U˙\dot{U} of xx also contains an infinite subset of SS, hence, xx is a limit point of SS.


Examples of locally finite sets (i.e. sets having no limit point):

1) any finite set;

2) the set of integers ZR\mathbb{Z}\subset\mathbb{R} (any open interval of length ll contains [l][l] or less integers);

3) the set S={1n:nN}S=\{\frac{1}{n}:n\in\mathbb{N}\} in the metric space X=(0,+)X=(0,+\infty) with the standard metric. Indeed, if x=1nSx=\frac{1}{n}\in S then the interval U=(1n+1,1n1)U=(\frac{1}{n+1}, \frac{1}{n-1}) is a neighborhood of xx such that the puncted UU contains no elements of SS. Therefore, xSx\notin S'. If xSx\notin S and x>0x>0 then put n=[1/x]n=[1/x].We have n<1/x<n+1n<1/x<n+1, thus, 1n+1<x<1n\frac{1}{n+1}<x<\frac{1}{n}. The interval U=(1n+1,1n)U=(\frac{1}{n+1}, \frac{1}{n}) is an open neighborhood of xx containing no elements of SS. Thus, xSx\notin S' and therefore, S=S'=\emptyset.


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