6

Exercise 2.8: Is every point of every open set $E\subset R^2$ a limit point of $E$ ?

My Solution:

Every point of every open set $E\subset R^2$ is a limit point of $E$.

[Notation: $N_r(p)$ is the set of all point x such that $0< d(x,p)< r $ ]

Since $E$ is open, let point $x \in E$, then x is an interior point of E. There is $r>0$ such that the deleted neighborhood $N_r(x) \subset E$. For any $s>0$, the deleted neighborhood $N_s(x)$ contains a point $z\in E$, if $0<d(x,z)<min(s,r)$. Thus $x$ is a limit point of $E$.

My question: I think my solution did not use any property of $R^2$, so this conclusion should be true in other metric spaces. Is that right?

Xianglin Zhao
  • 73
  • 2
    It's not true of the integers with the usual metric, so it appears you did use a property of $\mathbb{R}^2$. Where does your proof break down in you try it on $\mathbb{Z}$? – saulspatz Jul 13 '19 at 19:12

1 Answers1

5

Yes, you did use some property of $\Bbb R^2$ not shared by a general metric space. The fact that each deleted neighborhood $N_r(p)$ is not empty for every point $p\in \Bbb R^2$ is that property.

Counterexample to your conjecture can be easily found, e.g. if $(X,d)$ is a discrete metric space, then $N_{1/2}(x)$ is empty for every $x\in X$.

Ps. The fact that $N_r(p)$ is not empty for any $p$ is also valid for some class of spaces other than $\Bbb R^2$, e.g. any normed linear space. This includes $\Bbb R^n$ for all $n\in\Bbb N$.

BigbearZzz
  • 15,084