In terms of definition, I know the difference between the two. However, the set of rationals $\mathbb{Q}$ has measure zero but is dense in $\mathbb{R}$. Whenever I envision this, I see a set of measure zero to have gaps in it that stop it from having a full interval. At the same time, these gaps have length zero (so in a sense almost no gap) since the gaps are the irrationals and we can approximate any irrational with a rational number. Is there a better way to picture a set of measure zero than this since I seem to be chaining the concepts of density and measure too closely (or at least according to my instructor)
1 Answer
Ok this question is tricky!
As you pointed out that because $\mathbb{Q}$ is dense in $\mathbb{R}$ we will not find any open Set that contains just one rational number. As we will not find any open set that will contain just one irrational number.
"I see a set of measure zero to have gaps in it that stop it from having a full interval."
This is basically wrong because you might find sets like $\mathbb{R}\setminus \mathbb{Q}$ which contains also no interval.
But you can find a countable series of intervals $A_i$ for any $\epsilon >0$ so $\bigcup_{i=1}^\infty A_i \supset \mathbb{Q}$ and $\sum_{i=1}^\infty \mu(A_i) \le \epsilon$.
This is to say you can approximate $\mathbb{Q}$ by intervalls with arbitrarily small measure.
You will not be able to do this with $\mathbb{R}\setminus\mathbb{Q}$. As you would need way to much small Intervalls for them to have an arbitrarily small measure.
"I seem to be chaining the concepts of density and measure too closely"
Yep! A dense set need not have measure zero! As you have seen with $\mathbb{R}\setminus\mathbb{Q}$. Obviously also not all sets with measure zero are dense!
-
$\begingroup$ Great answer! The picture in my head is a bit clearer now $\endgroup$ Feb 26, 2014 at 0:09