I've been asked to give an example of a closed subset of a metric space that cannot be expressed as an intersection of closed balls. Any ideas?
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What ideas come to your mind? The first thing I thought of worked, so it isn't meant to be a challenging problem, just a conceptual one. – Clayton Feb 24 '16 at 23:27
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I just can't seem to think of anything... Can the whole of a set be expressed as an intersection? – Lucy Keen Feb 24 '16 at 23:55
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Hint: In $R^2$, consider the line $x=0$.
In fact take any closed and unbounded subset of $R^n$.
Tsemo Aristide
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Thank you for your help, but how can a set of finite points be closed? – Lucy Keen Feb 24 '16 at 23:55
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@LucyKeen Finite sets are closed in $\mathbb{R}^n$, since they can't have any limit points, hence vacuously closed. – Ian Feb 25 '16 at 00:18
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May not be true for general metric space. Consider a set has only one point. This point is only closed set that is closed (also open). The point is an intersection of the closed ball.
The question may need extra condition. For example, in Euclidean space, any two points cannot be the intersection of the closed balls, since there are many points on the line segments of two points.
On Euclidean space with the Euclidean distance, if two separate points are the intersections of closed balls, those two points must be in each clsoed ball. Since it is the ball, the line segment connecting those two points are in in the closed ball. Hence the line segment in every closed ball, that is a contradiction.
runaround
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@lhf, the intersection of all of them is empty set, right? 0 is not in the ball around 1, and 1 is not around the ball around 0. – runaround Feb 25 '16 at 01:27