- #1

- 147

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ForMyThunder
- Start date

- #1

- 147

- 0

- #2

- 22,089

- 3,297

But for every open set that covers the rationals in, say, [tex] [0,1] [/tex] must cover the entire interval

This is where your argument goes wrong. There are many open sets around the rationals that does not cover the entire set. For example, the set [itex]\mathbb{R}\setminus \{\pi\}[/itex] cover the rationals but not the entire real line.

In fact, since the rationals are countable, we can enumerate them as

[tex]\mathbb{Q}=\{q_1,q_2,q_3,...\}[/tex]

then

[tex]G_\varepsilon=\bigcup_{n\geq 1}{B(q_n,\varepsilon/2^{n+1})}[/tex]

is an open set containing the rationals and the open set has measure epsilon. Fine, I have no inuition about that open set and I can't imagine it, but that doesn't mean it doesn't exist!

- #3

disregardthat

Science Advisor

- 1,861

- 34

[tex]G_\varepsilon=\bigcup_{n\geq 1}{B(q_n,\varepsilon/2^{n+1})}[/tex]

is an open set containing the rationals and the open set has measure epsilon.

Hm, I can't see how that is correct. First of all, [itex]\sum_{n \geq 1} \frac{1}{2^{n+1}} = 1/2[/itex], and second of all, it is not a disjoint union. Also, I think the measure would depend on the how Q is ordered.

- #4

- 22,089

- 3,297

Hm, I can't see how that is correct. First of all, [itex]\sum_{n \geq 1} \frac{1}{2^{n+1}} = 1/2[/itex], and second of all, it is not a disjoint union. Also, I think the measure would depend on the how Q is ordered.

Sorry, that should be "measure at most epsilon", which is enough for our purposes: Thus

[tex]\lambda\left( \bigcup_{n\geq 1}{B(q_n,\epsilon/2^{n+1})} \right)\leq \sum_{n\geq 1}{\lambda(B(q_n,\epsilon/2^{n+1}))}=\sum_{n\geq 1}{\frac{\epsilon}{2^n}}=\epsilon[/tex]

And of course the measure of the set, and the set itself will depend on the order of Q! For every different order, we get a different set!

- #5

disregardthat

Science Advisor

- 1,861

- 34

And of course the measure of the set, and the set itself will depend on the order of Q! For every different order, we get a different set!

Not necessarily though, q_1,q_2,q_3,q_4,q_5,... and q_1,q_3,q_2,q_4,q_5,... would generate the same set if the balls around q_2 and q_3 are contained in the ball around q_1.

And furthermore, different sets can easily have the same measure! (not saying that you implied otherwise..) As an example, relabeling by reflection about 1/2 would yield a (potentially) different set with equal measure.

But this example is interesting regardless, it is quite difficult to conceive of a set containing small intervals around each point in a dense set while not containing its closure.

Share:

- Replies
- 8

- Views
- 10K