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In general metric spaces the closed ball is not the closure of an open ball.

However, I read that in the Euclidean space with usual metric, closed ball is the closure of an open ball. I'm having trouble rigorously proving this? How can I show this?

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    $\begingroup$ Show that $\{x : \|x\| \le 1\}$ is the closure of $\{x : \|x\| < 1\}$; try doing this first in $1$ or $2$ dimensions to get a feel for the general proof. $\endgroup$ – user61527 Mar 23 '14 at 18:04
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The only reason a closed ball might not be the closure of an open ball is when there is a "hole" between a point on the sphere and the centre.

Euclidean spaces have no "holes": if you take a point $x$ on a sphere of radius $r$ centred at a point $x_0$, then there is an interval connecting $x_0$ to $x$ and except $x$ it lies within an open ball of radius $r$.

The same argument works for normed spaces and general geodesic spaces, and with slight modification, for arbitrary length metric spaces.

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Hint:

Suppose $B_x(r)$ and $B_x[r]$ are the respectively open and closed balls with centre $x$ and radius $r \gt 0$. If $y \in B_x[r]$ then $||y - x|| \le r \implies ||y - x|| \lt r$ or $||y - x|| = r$. For the first case since $y \in B_x(r)$ it is easy to prove that $y$ is also in the closure since $ A \subseteq A^{\circ} $.

For the second case, assume $y$ is not in some closed set $C$ which contains $B_x(r)$. Since $C$ is closed it contains its interior and boundary points and hence $ y $ is an exterior point for $C$ and hence for $ B_x(r) $. Now consider any neighbourhood of $y$.

Hence try to prove that $y$ is in every closed set which contains $ B_x(r) $ and you would be done.

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  • $\begingroup$ I tried to prove it using your hint. The first case is obvious and for the second case, I think you are trying to prove it by contradiction. Following your hint, I used the fact that Ext($B_x(r)$)=Int($X\B_x(r)$) and therefore y is an interior point of $X\B_x(r)$, which means that theres an open ball that does not intersect with $B_x(r)$. And now this contradicts the assumption that ||y−x||=r. $\endgroup$ – nomadicmathematician Mar 23 '14 at 18:30
  • $\begingroup$ But following this proof, I don't quite get which part makes it insufficient to make this proof be generalized to the case of all metric spaces. Could you tell me why this proof can only be applied to the Euclidean space with usual metric? $\endgroup$ – nomadicmathematician Mar 23 '14 at 18:34
  • $\begingroup$ The argument for the proof looks good. Well first of all I've been learning Analysis only on the Euclidean space and my knowledge on general metric spaces is not great. Either way I think your question has been sufficiently answered by the second response below. $\endgroup$ – Ishfaaq Mar 24 '14 at 1:15

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