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I have read that the measure of the irrational numbers on an interval $[a,b] = b-a$. This both makes sense and doesn't make sense to me. If you consider that the union of the irrationals with the rationals are the reals, then if the rationals have measure 0, then the irrationals must have the same measure as the reals. (right?)

But if you consider that the irrationals are totally disconnected, then it is a collection of disconnected points. If it's a collection of disconnected points, like the rationals, how can it have a non-zero measure?

Another fact I'm not sold on, which may be the source of the problem, is that there isn't a bijection between the rationals and the irrationals. If both of these sets are totally disconnected by each other and both dense, that is, "Between any $ \forall a,b \in \mathbb{R} $, such that $a < b$, there is exists an irrational $i$ and a rational $q$ satisfying $a < i < b$, and $a < q < b$."

I know of the diagonalization argument but I've just never been completely sold on this fact. For the irrationals to be uncountable and the rationals to be countable, in my head it would make more sense if there exists an $\epsilon > 0$ such that around any irrational number there exists only other irrational numbers. I know that if it exists you could just sum enough of those epsilons together to get the length of it greater than some $\frac{p}{q}$ thus finding a contradiction... but I'm still not sold on it.

Can someone help me finally understand these concepts?

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    $\begingroup$ "If it's a collection of disconnected points, like the rationals, how can it have a non-zero measure?" Because it is an uncountable collection, and things get strange when you leave the countable world. $\endgroup$ – uniquesolution Nov 18 '15 at 14:17
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    $\begingroup$ Welcome to the wild and wacky jungle of mathematical analysis! Stick with it, it all becomes clear in time. $\endgroup$ – icurays1 Nov 18 '15 at 14:19
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    $\begingroup$ You are trying to make your intuition precede a serious study of these topics. Work with the ideas, study them, solve problems, read some history (perhaps about Peano-Jordan measure, Borel's ideas, why Lebesgue measure was introduced, etc.) Finally your intuition develops and they "make sense." No sane person, first encountering these ideas, feels that they fit neatly into the way that he thinks about "numbers" and "measure" and "counting" and "infinite." Rely on the definitions, the theorems, hard work on problems, to help you develop a picture of how it all works. $\endgroup$ – B. S. Thomson Nov 18 '15 at 17:14
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    $\begingroup$ Well, one problem I can spot is that you are throwing around a jumble of terms and ideas and can't sort them out: you have apparently studied what "totally disconnected set" is but then interpreted this to mean that somehow this is a bunch "disconnected points." I'm sure nobody has given you a definition for "disconnected points." This is your intuition running wild. Similarly the phrase "totally disconnected by each other." I did a search and in the entire history of google you are the sole person to use this phrase. You are inventing ideas (not a bad thing) and then getting confused. $\endgroup$ – B. S. Thomson Nov 18 '15 at 21:52
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    $\begingroup$ The reason the rationals have measure zero is not because they are "disconnected." $\endgroup$ – Thomas Andrews Nov 19 '15 at 5:28
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The set of rational number is numerable thus $Q=(u_n)_{n\in N}$. Each rational is a Borelian set of measure 0, thus $\mu(u_n)=0$, $\mu(Q)=lim_{n>0}\mu(U_n)$ where $U_n=\{u_0,...,u_n\}$. Thus $\mu(U_n)=0$ and henceforth $\mu(Q)=0$. Since $R$ is the disjoint union of $Q$ and $R-Q$, and $\mu(Q)=0$, then $\mu(R-Q)=+\infty$.

You also have $[a,b]\cap Q\subset Q$ thus $\mu([a,b]\cap Q)=0$. $[a,b]$ is the disjoint union of $[a,b]\cap Q$ and $[a,b]-Q$ thus $\mu([a,b])=\mu([a,b]\cap Q)+\mu([a,b]-Q)=b-a$.

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    $\begingroup$ But the OP is asking about irrational numbers. $\endgroup$ – mrp Nov 18 '15 at 14:54
  • $\begingroup$ @mrp Didn't he already prove that $\mu(\mathbb{R}-\mathbb{Q}) = +\infty$ where $\mathbb{R}-\mathbb{Q}$ the irrationals? $\endgroup$ – ares Oct 12 '18 at 2:09

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