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According to Wikipedia, the straight line $\mathbb{R}$ is a null set in $\mathbb{R}^2$.

That means, the line $\mathbb{R}$ can be contained in $\bigcup_{k=1}^\infty B_k$, where $B_k$ are open disks and their total measure of all the $B_k$ is less than $\epsilon$.

Question 1: How can this be done? Any explicit construction to show this?

Question 2: Since the intersection of $B_k$ with $\mathbb{R}$ is an open interval $I_k$, doesn't this mean that $\mathbb{R}$ can be covered by union of intervals $I_k$ whose total length is arbitrarily small? (Which according to my previous question is impossible?)

Sincere thanks for any help. I am puzzled by this.

Community
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yoyostein
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2 Answers2

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First, about Question 2, a 2-dimensional null set doesn't have to be an 1-dimensional null set.

Answering Question 1: You proceed as follows:

  1. Easily show that any interval $[n,n+1]$, $n\in\mathbb{Z}$ is a 2-dimensional null set.
  2. Use the following result from the real analysis: An enumerable union of null sets is still a null set.
  3. Conclude your statement by writing $\mathbb{R}=\cup_{n\in\mathbb{Z}} [n,n+1]$ and using 2 and 3.
ACV
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Let me address one subtlety that ACV did not. For each $\epsilon$, you can find balls $B_k$ such that they cover the line in the plane, and furthermore, $$ \sum_{k = 0}^\infty V(B_k) < \epsilon. $$ (I use $V$ for the measure.) Then you can take the intersection of each ball $B_k$ with the real line to get an interval $I_k$, and intuitively it seems that also $$ \sum_{k = 0}^\infty V(I_k) < \epsilon. $$ However, if $V$ here means the measure on $\mathbb R$, then this is not true. The reason is, basically, that the surface of a ball with radius $\delta$ is something like $\pi\delta^2$, while the length of an interval with "radius" $\delta$ is $2\delta$. And $\delta^2\pi$ tends to 0 much faster than $2\delta$ if $\delta$ tends to 0. So, you can find infinitely many balls that are sufficiently small to cover the line, but for intervals you will never reach "sufficiently small".

This also gives you a hint about how to construct the radiuses of the $B_k$, if you want do do so explicitly.

(This generalizes to what I am pretty sure is called Hausdorff dimension, if you are interested.)

Mees de Vries
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  • So let the radii be like $\epsilon/n$ and hence of infinite length in total but tiny area. I think your explanation is a bit thick, but adds important points. We don't need to break it down into a union of intervals but can construct it directly and easily. – jdods May 11 '16 at 13:09