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The boundary of a subset of Euclidean space has empty interior, and furthermore has Lebesgue measure zero.Well,this is generally not true,but I can't find an explicit counter-example right now.

Similarly,in set topology.Please show that there exists a metric space,such that the closure of the open ball of radius r, is not the closed ball of radius r in that metric space

Motivation:Actually,there are many such kinds of examples in measure theory and set topology which are not consistent with our intuitions.I find that clarifying these fuzzy definitions or false believes may sometimes be very helpfui.

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$\mathbb{Q ,Z}$ –  Nate Eldredge Sep 1 '12 at 15:34

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An example of a subset of $\mathbb{R}$ whose boundary has positive Lebesgue measure is a "fat Cantor set". Mimic the usual construction of Cantor's middle-thirds set, but at the $n$th stage remove, say, the middle $(1/4)^n$ from each of the remaining sub-intervals. This will result in an totally disconnected perfect set of positive Lebesgue measure. As it is its own boundary, its boundary has positive Lebesgue measure. (Of course, as it is nowhere dense, its boundary still has empty interior.)

To get a subset of $\mathbb{R}$ whose boundary has nonempty interior, this set must have the property that both it and its complement is dense in some open interval. The rationals (or the irrationals) fit this purpose quite well.

A simple example of a metric space in which the closure of a ball with radius $r$ is not the closed ball of radius $r$ is to take the discrete metric on any set $X$ (with at least two elements): $$d(x,y) = \begin{cases} 0, &\text{if }x = y\\ 1, &\text{if }x \neq y.\end{cases}$$ Then for $x \in X$ we have that $\overline{B_1 (x)} = \overline{\{ x \}} = \{ x \} \neq X = \overline{B_1} (x)$.

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