Let $(X,d)$ be a metric space, $x\in X$ and $r>0$. The set $$B(x,r)=\{y\in X: d(x,y)<r \}$$ is called an open ball centered at $x$ and radius $r$.
In the usual metric space, let $x=1$, then $(0.5, 1.5)$ is an open ball of radius $r=0.5$ centered at $x=1$.
Now, I want to show that $(0.5, 1.5)\cup (5,7)$ is not an open ball of radius $r>0$ centered at $x=1$.
My steps are:
- First, an open ball of radius $r>0$ centered at $x=1$ is an open interval $(1-r,1+r)$.
- Then I need show $(0.5, 1.5)\cup (5,7)\neq (1-r,1+r)$. To do that, I consider some cases, i.e.
(i) if $0<r<1$, then there exists $6\in (0.5, 1.5)\cup (5,7)$, but $6\notin (1-r,1+r)$.
(ii) if $r=1$,then $(1-r,1+r)=(0,2)\nsubseteq (0.5, 1.5)\cup (5,7)$.
(iii) if $r>1$ then there exists $1-\frac{r}{2}\le 0$ such that $1-\frac{r}{2}\in (1-r,1+r)$ but $1-\frac{r}{2}\notin (0.5, 1.5)\cup (5,7)$.
From (i)-(iii), I get $(0.5, 1.5)\cup (5,7)\neq (1-r,1+r)$.
Are my steps above correct?
Thanks for any help.
