Correct Measurable subgroups of $\mathbb R$

If $(\mathbb{R},+)$ is a team and also $H$ is a correct subgroup of $\mathbb{R}$ after that confirm that $H$ is of action absolutely no.

0
2019-05-05 00:48:38
Source Share
Answers: 2

Your title speak about "correct quantifiable subgroups of R", yet the body of your blog post does not call for that H be quantifiable. The adhering to synopsis reveals that if H is a correct subgroup of R and also is quantifiable, after that it has to be action 0. I'm not exactly sure if every subgroup has to be quantifiable or otherwise

Here is a harsh synopsis of the evidence :

Lemma 1 : If H is a part of R and also has favorable action, after that H - H = a, b, in H has a period around 0. Evidence : See http://unapologetic.wordpress.com/2010/04/23/lebesgue-measurable-sets/

For Lemma 2 and also 3, I'll leave the evidence to you (yet I can add information if you require them).

Lemma 2 : If H is a subgroup of G, after that H - H = H.

Lemma 3 : If H is a subgroup of the actual numbers R and also has a period around 0, after that H = R.

0
2019-05-08 22:26:05
Source

As was kept in mind by Jason DeVito if H is quantifiable after that action of H is 0.

From the various other hand, if we intend, that the axiom of selection holds, there is an opportunity, that H is not quantifiable. The evidence is fairly straightforward. $\mathbb{R}$ is a vector room over $\mathbb{Q}$ . Consequently there is some basis. Intend e is just one of it's components and also H is a subspace in $\mathbb{R}$ , created by others. After that H is a subgroup of ( $\mathbb{R}$ ,+).

Lemma : H is not quantifiable.

Intend H is quantifiable. After that as was kept in mind over, it's action m (H) = 0. After that every set of the kind $H+q e = \{h+qe, h\in H\}$ , where $q\in Q$ , has action 0 (due to the fact that it is simply a change of H). Yet $\mathbb{R}$ amounts to union of countable several collections with action 0 : $\mathbb{R} = \cup_{q\in\mathbb{Q}}(H+qe)$ . Consequently $m(\mathbb{R})=0$ . We have actually involved an opposition.

Additionally there is straightforward evidence of the reality, that if H is quantifiable, after that H is of action 0.

Lemma : If H is quantifiable correct subgroup of $\mathbb{R}$ , after that m (H) = 0.

If H = 0 after that suggestion of the lemma is noticeable. Or else we can locate favorable component z in H. Suppose, $H_0 = H\cap [0,z)$ . If $m(H_0)=0$ after that $m(H)=0$ . Or else $m(H_0)=\delta>0$ . Allow's take integer N, such that $\delta N > z+1$ (we will certainly see later on, why). Keep in mind that if x is not in H, after that x/n (for every single favorable integer n) and also - x are additionally not in H. Therefore, making use of the reality that H appertains, we can locate favorable x < 1, such that $x\notin H$ . Intend y = x/N!. After that for $n=1,\dots,N$ number ny complies with the adhering to buildings :
1. 1 > ny > 0.
2. ny is not in H.
After that establishes $H_0, H_0+y, \dots, H_0+(N-1)y$ are disjoint parts of $[0, 1+z)$ . Consequently, $\displaystyle 1+z = m\Big( [0, 1+z) \Big) \geq m\Bigg(\bigcup_{n=0}^{N-1} (H_0 + n y)\Bigg) = N \delta.$ Here we have an opposition with definition of N.

0
2019-05-08 19:19:47
Source