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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1How about the union of the Cantor set and the set of all rational numbers? – bof Aug 23 '14 at 06:28
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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. – Andrés E. Caicedo Aug 23 '14 at 06:32
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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.
Chris Eagle
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I say not compact Your set is compact beacause cantor set is compact set – kaveh Jul 26 '12 at 18:23
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@kaveh: That is not true in general, $(0,1]$ is not compact, but we only removed ${0}$ from $[0,1]$. – Asaf Karagila Jul 26 '12 at 19:12
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While this answers the question as stated, I prefer the answer of @AsafKaragila, since it satisfies the much more interesting property that any neighbourhood of any point in his set contains a boundary point not in the set. So it is much farther from being closed. – Harald Hanche-Olsen Jul 26 '12 at 19:23
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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.
Martin Sleziak
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Asaf Karagila
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why The intersection is still non-empty, but the result is something homeomorphic to the irrational ? – kaveh Jul 26 '12 at 18:45
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We intersect open sets which contain compact sets, so the intersection is non-empty by the same argument as in the Cantor set. The result is not closed, that is the main point here. The fact this is homeomorphic to the irrationals is irrelevant. – Asaf Karagila Jul 26 '12 at 18:53
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1You 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. – Harald Hanche-Olsen Jul 26 '12 at 19:19
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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.
André Nicolas
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I changed the wording to make it clearer. If $C$ is the Cantor set, let $B$ be the collection of all points of the form $n+c$, where $n$ ranges over the integers and $c$ ranges over $C$. – André Nicolas Jul 26 '12 at 18:31
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Take the union of Natural numbers with Cantor set which is an unbounded , uncountable set of Labesgue measure zero.
GA316
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