Uncountable measure zero subset

1. Feb 25, 2009

Kreizhn

1. The problem statement, all variables and given/known data
Construct a sequence $(G_n)$ of dense open subsets of $\mathbb{R}$ such that
$$\lambda \left( \bigcap_{n=1}^\infty G_n \right) = 0$$
and hence deduce that there exists an uncountable subset of $\mathbb{R}$ of measure 0

2. Relevant equations
$\lambda$ is the Lebesgue measure on $\mathbb{R}$

3. The attempt at a solution
This really just seems like a pseudo-extension of the Cantor set to $\mathbb{R}$, but I'm not sure how to go about constructing it.

2. Feb 25, 2009

Dick

Uh, forgive me if I'm being dense here. But let q:Z->Q be a bijection of the integers to the rationals. Let G_n be the union of the open intervals (q(k)-1/2^(n+k),q(k)+1/2^(n+k)). Isn't each G_n open and dense? Isn't the intersection of all the G_n just the rationals? Hence having measure zero? How does this help you to prove the existence of an uncountable set of measure zero? I will agree the Cantor set is an example of what you want. But I don't see how this sequence of open dense sets will get you there. Am I missing something?

3. Feb 25, 2009

Kreizhn

Pun intended?

Thank you for this example, it was quite useful. I believe that this gives us an uncountable subset of $\mathbb{R}$ of measure zero, since if a $G_\delta$ subset of $\mathbb{R}$ is dense, it can't be countable. Hence this example gives a dense $G_\delta$ set, and we are forced to conclude that it is uncountable.

4. Feb 25, 2009

Dick

Pun intended. But I also think I might be missing something. I tried to give you an example of a set where the intersection of the open dense sets G_n is, in fact, countable. I don't see how this strategy leads you to conclude the existence of an uncountable set of measure zero.

5. Feb 25, 2009

Kreizhn

Good point. Assuming that my result about the uncountability of Gdelta sets is true (and it might not be) then could it be that the intersection is not necessarily that rationals? The Gn are open since they're the union of open sets, and dense since the rationals are contained within. It seems like a problem (if there is one) might occur in the intersection, though I'm not sure how to resolve it immediately.

Edit: I don't think the intersection could be the rationals, otherwise the example would give a G_delta construction of the rationals, which is impossibly by Baire's Theorem.

6. Feb 26, 2009

Kreizhn

Nonetheless, I think that your example is correct, since of $G = \cap G_n$ then $G \subset G_n, \forall n$ and so $\lambda(G) \leq \lambda(G_n), \forall n$ and the measure of each $G_n \to 0 \text{ as } n \to \infty$

7. Feb 26, 2009

Dick

I think you are right. I'd better review Baire category stuff.

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