4

Definition of Looped line At each point $x$ of the real line other than the origin, the basic neighborhoods of $x$ will be the usual open intervals centered at $x$. Basic neighborhoods of the origin will be the sets

$(-\epsilon,\epsilon)\cup(-\infty,-n)\cup(n,\infty)$ for all possible choices $\epsilon >0$ and $n \in \mathbb{N}$. This gives a topology on the line.

Problem. Prove the looped line is Hausdorff.

I know, by definition of Hausdorff space, with different points $x,y$ and differents of zero, we can disjoint open. But when, for example, $x=0$ there's no exists a open of $y$ such that both are disjoints. Could you guide me to solve this problem?

PSW
  • 167
  • 1
    Consider $0$ and $\pi$, for instance: the nbhd $(-1,1)\cup(-\infty,-5)\cup(5,\infty)$ of $0$ and the nbhd $(3,4)$ of $\pi$ are disjoint. You can do something similar with $0$ and any other number. – Brian M. Scott Apr 19 '20 at 23:25

1 Answers1

1

For $x = 0$ and $y\neq 0$ WLOG suppose that $y \gt 0$, consider the interval

$(\frac{y}{2},\frac{3y}{2})$ as an nhood for $y$ and

$(\frac{-y}{2},\frac{y}{2}) \cup (-\infty,-\lceil\frac{3y}{2}\rceil) \cup (\lceil\frac{3y}{2}\rceil,\infty)$ as an nhood of $x$.

Clearly these two nhoods are disjoint and based on the definition of nhood, you can find open sets $V_y$ and $U_x$ contained in $y$-nhood and $x$-nhood respectively, so $U_x$ and $V_y$ are disjoint.

For $y \lt 0$ you can do the same.

schneiderlog
  • 162