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In other words does there exist a non discrete metric space in which Closed balls are subset of the closure of open balls?

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4 Answers 4

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None of the explicit examples given so far seem to be important as anything other than "weird counterexamples" (similar to the examples one may create to test ideas in general topology). It would be good to have an answer to the question using a metric space that is of serious interest. One such example is the $p$-adic numbers ${\mathbf Q}_p$, which form a metric space in which any open ball is already a closed set. For example, the open unit ball $\{x \in {\mathbf Q}_p : |x|_p < 1\}$ is a closed set, so its closure is itself. The unit circle $\{x \in {\mathbf Q}_p : |x|_p = 1\}$ can't be approached closely by any point in the open unit ball. Another example is the $X$-adic metric on the formal power series ring ${\mathbf R}[[X]]$, which can be used to justify some formal manipulations with formal power series by checking the case of polynomials and then appealing to continuity.

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enter image description here

The rectangle pictured above, with the Euclidean metric restricted from the plane, gives an example where the closure of an open ball of radius $r$ is not the same as the closed ball of radius $r$. The segment inside the pictured circle (not including endpoints) is an open ball, and its closure is the segment including the endpoints. However, the closed ball with the same radius includes the point where the circle is tangent to the rectangle.

Because closed balls are closed, the closure of an open ball of radius $r$ is always contained in the closed ball of radius $r$, so the only way for the two to not be equal is for the closed ball to contain points not in the closure of the open ball.

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3  
To be explicit: by rectangle you mean the green segments shown, not the interior. –  Ross Millikan Nov 5 '11 at 16:46
    
Ross: That's right, thanks. –  Jonas Meyer Nov 5 '11 at 16:51

To construct an example as in the title (the text doesn't say the same thing): remove from $\mathbb R^2$ the intersection of the open unit disk and the rectangle $[1/2, 1]\times [-1,1]$. In the new space (with the induced metric), the closure of the open unit disk is strictly smaller than then the closed unit disk (we miss a portion of the circle).

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right ..thank you very much –  Abcd J Nov 5 '11 at 18:14

Maybe the OT would mean: the closure of the open ball is a subset of the closed ball (not the converse). Here is another simple example: $(\mathbb R,d)$, with $d(x,y)=\min\{1,|x-y|\}$.

The condition that the closed balls are the same as the closure of the balls is known to be equivalent to the following condition: for all $x\in X$, the mapping $y\rightarrow d(x,y)$ has a unique propert minimum at $x$. This is why metric spaces verifying this property are called pitless in http://arxiv.org/abs/1111.0268. They verify also a general property, which generalizes amenability to any metric space, called property SN.

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