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Topic: Measure and Density
Replies: 14   Last Post: Feb 23, 2013 11:26 AM

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quasi

Posts: 10,257
Registered: 7/15/05
Re: Measure and Density
Posted: Feb 16, 2013 10:56 PM
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quasi wrote:
>William Elliot wrote:
>>
>> [User "Herb" on forum "Ask An Analyst" asked]:
>>
>>How can we find a measurable dense subset S of [0,1], with
>>m(S) < 1, and such that for any (a,b) in [0,1], we have
>>m(S /\ (a,b)) > 0?

>
>Let Q denote the set of rational numbers and let
>
> x_1, x_2, x_3, ...
>
>be an enumeration of Q /\ (0,1).
>
>For each positive integer k, let
>
> a_k = max(0,x_k - 1/(2^(k+1)))
>
> b_k = min(1,x_k + 1/(2^(k+1)))
>
>and define the open interval I_k by
>
> I_k = (a_k,b_k)
>
>Finally, let S be the union of the intervals
>
> I_1, I_2, I_3, ...
>
>Then S satisfies the required conditions.


Now that I see Rotwang's solution (our solutions are
essentially the same with a minor difference), I realize
that I should have used

a_k = max(0,x_k - 1/(2^(k+2)))

b_k = min(1,x_k + 1/(2^(k+2)))

to make it more evident that m(S) < 1.

However, my original solution was not actually wrong since, for
any enumeration of Q /\ (0,1), the intervals

I_1, I_2, I_3, ...

cannot be pairwise disjoint, hence it's still true that m(S) < 1.

quasi



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