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I have been asked this question, but I do not understand it.

"For what values of $\alpha$ and $\beta$ are the sets $(\alpha, \beta)$, $[\alpha, \beta)$, $(\alpha, \beta]$ and $[\alpha, \beta]$ open balls in the metric space $[a, b]$?"

We have our interval $[a,b]$ which can be interpreted as a line along the x-axis. How can, for example, $(\alpha, \beta)$ which is another interval along the x-axis be considered a ball if it only exists on the x-axis and doesn't extend to the y-axis?

dustin
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frightenedeyes
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    An open ball (with respect to the metric $d$) is by definition a set of the form $$B_r(x) = { y : d(x,y) < r}$$ for some $r > 0$. Since in $\mathbb{R}^3$ with the Euclidean metric, these things are balls, the name has been generalised to arbitrary metric spaces. In $\mathbb{R}$, open balls are open intervals, in $\mathbb{R}^2$, open disks (all with the Euclidean metric). – Daniel Fischer Nov 07 '14 at 20:42

2 Answers2

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Short answer - notation convention.

Long answer - the ball in $\Bbb R^n$ is a set $$B(x,r) = \{y\in \Bbb R^n:\,|x-y|<r\},$$ i.e. it is centered in $x$ and has the radius $r$.

In the case $n=1$ the ball degenerates to the interval $(x-r,x+r)$.

TZakrevskiy
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  • Thank you. So for the first part of the question with the interval(α, β) we can let α and β equal anything in [a,b] so long as α≤β ? – frightenedeyes Nov 07 '14 at 21:03
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Balls aren't always circles or spheres. Consider $(x-\epsilon,x+\epsilon)$. It is a ball centered at $x$ and radius $\epsilon$. It is an open ball by definition. (You can view it as the intersection of a circle (or sphere) of radius $\epsilon$ centered at $x$ in $\mathbb{R}^2$ (or $\mathbb{R}^3)$,and its itersection with $\mathbb{R}$)

Swapnil Tripathi
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