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If we write $B(0,\infty)$ as the open ball then $\mathbb{R}$ is an open ball in $\mathbb{R}$. Is it correct?

Kamil Jarosz
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uuuuuuuuuu
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    The radius of a ball should be a positive real number. – gniourf_gniourf Feb 11 '16 at 16:03
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    However, using the standard bounded metric $\hat{d}(x,y)=\min{d(x,y),1}$, one does have that $\mathbb R = B_{\hat{d}}(0,2)$. This metric induces the same topology on $\mathbb R$ as the usual euclidean metric $d$. So $\mathbb R$ is an open ball in the metric space $(\mathbb R,\hat{d})$. Moral: you must specify a metric to talk about balls (though I understand your implicit metric). – MPW Feb 11 '16 at 16:08
  • yes, the metric is the usual euclidean metric i.e., distance between two points in $\mathbb{R}$. – uuuuuuuuuu Feb 11 '16 at 16:24
  • Is it true that if two metrics are topologically equivalent then then an open ball w.r.t one metric is also an open ball w.r.t another metric and vice-versa. – uuuuuuuuuu Feb 11 '16 at 16:47

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No, balls have finite radius. It is an open set however.

Stella Biderman
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    It is also closed. – trying Feb 11 '16 at 16:07
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    That's usually what the word "no" means. A ball has two determining features: a center point, and a radius distance. The radius is usually a positive real number. I don't know of any text or person who allows the radius to be infinite. I suppose one could but than one will have to always make specifications for finite balls and I can't see anything gained in making the reals a ball. What you can't (definitely never) is define a ball with zero radius. That would be the empty set and it'd make every set open. – fleablood Feb 11 '16 at 16:14