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I find it difficult to understand why the 'size' of the set of rational numbers in an interval such as [0,1] is zero. I know that there are way more irrational numbers than rational numbers such that m(set of irrational numbers) = 1 and as such m(set of rational numbers)=0.

But I still find it difficult to reconcile with the fact that there are infinitely many rational numbers in the interval [0,1].

Can someone provide an explanation for this?

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Because of the fact that $$m([0,1] ) = m(\mathbb{Q} \cap [0,1]) + m(\mathbb{Q}^c \cap [0,1]). $$ –  Nameless Aug 7 at 3:26

2 Answers 2

up vote 4 down vote accepted

The definition of measure means that if $(X_i)_{i\in\mathbb N}$ are measurable and disjoint, then $$\mu(\bigcup_i X_i) = \sum_i \mu(X_i)$$ In particular, if each $X_i=\{x_i\}$ is a single point, then $\mu(X_i)=0$ and so $$\mu(\bigcup_i X_i) =0$$

Alternatively, let $X=\{x_1,x_2,\dots\}$ be a countable set of real numbers, and, for $\epsilon>0$, let $I_1,I_2,\dots$ be a sequence of open intervals, with $x_i\in I_i$ and $\mu(I_i)=\epsilon/2^{i+1}$. Then the total measure of $\bigcup_i I_i$ is at most $\frac{\epsilon}2$. That means that $\mu(X)<\epsilon$ for evry $\epsilon>0$, which means $\mu(X)=0$.

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I would first ask whether it is the infinity of the set or the density that is throwing you off. If it is truly the infinite nature of the set, perhaps it might be better to play with the Cantor set first. In fact, this may seem even more paradoxical in that case, as the Cantor set has the same cardinality as $[0,1]$. But it is more evident here without density obscuring things that we have measure zero. The cantor set can be contained by construction in $2^n$ intervals of length only $3^{-n}$ for all $n$, thus has measure less than $(2/3)^n$, so it seems fairly clear it must be measure zero.

If instead it is the density, I might recommed you tackle the perhaps more flummoxing notion that there are open and dense subsets of $[0,1]$ with arbitrarily small measure. In other words we have sets $U$ that are dense, also contain some small interval around every point of $U$, and yet somehow not only fail to cover every point, but fail to even cover many of them.

To see why this might be reasonable, consider something simple like $(0,1)$, now obviously this has full measure in $[0,1/\pi)\cup(1/\pi, 1]$, but it makes the point that an open set covering all rationals an missing a point, so the fact that a set is open about rationals certainly doesn't mean it covers everything, in a sense this is because if you look at numbers going towards $1/\pi$ the intervals we may find around them shrink somewhat quickly. If we impose some even more severe restrictions on interval size, we can get even better results.

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