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I want a example of a set that it is uncountable and has measure zero and not compact? Cantor set has these properties except not compactness.

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    An obvious solution is to note that any subset of the Cantor set has measure zero. There are $2^{\mathfrak c}$ subsets, and only $\mathfrak c$ of them are closed or countable, so $2^{\mathfrak c}$ subsets of the Cantor set serve as examples.2014-08-23

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Take the union of Natural numbers with Cantor set which is an unbounded , uncountable set of Labesgue measure zero.

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Just delete a point, say $0$, from the Cantor set and you'll get a set with the desired properties. In fact, since the Cantor set is perfect, any point will do.

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    yes I take it thanks2012-07-26
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Recall the construction of the Cantor set. At each step we remove open intervals, now remove closed intervals.

The intersection is still non-empty, but the result is something homeomorphic to the irrational numbers, or more generally, Baire space.

This is not a compact set, since we can show that the points removed are in the closure of this new set, and thus it is not closed. However as a subset of the Cantor set it still has measure zero.

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    You need also to argue that this set is not countable. However this is easy, since the difference between this set and the Cantor set is countable – it consists of the endpoints of the intervals removed. Corollary: The set is nonempty.2012-07-26
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Let $B$ be the union of translates of the Cantor set by every integer $n$. Then $B$ is uncountable, has measure $0$. It is unbounded so not compact.

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    thanks I understand this explain2012-07-26