# How to construct a dense subset of $\mathbb R$ other than rationals.

How to construct a dense subset say $A$, of the real numbers other than rationals? By dense I mean that there should be an element of $A$ between any two real numbers.

• How about $\mathbb R$? – Bruce George Mar 3 '12 at 15:49
• Add some irrational number - pi, for example - to every rational. – Mark Bennet Mar 3 '12 at 15:50
• How about $\mathbb{R} \setminus \mathbb{Q}$? – Andy Mar 3 '12 at 15:51
• An interesting question arises when you if there is a dense in $R$ measurable such that $m(A\cap I)=1/2 |I|$ for any interval this property also tell us that the complement is a set of the same kind. – checkmath Mar 3 '12 at 20:26
• If I understand your comment, that for any $I$, $m(A\cap I) = 1/2 |I|$, then this is impossible by the Lebesgue Density Theorem. – Patrick Mar 4 '12 at 1:49

First note that you can "cheat" by taking any subset and take a union with the rationals.

Second note that you can always cheat by taking some real number $x$ and considering the set $\{x+q\mid q\in\mathbb Q\}$. If $x$ is irrational then the set is not the rationals.

Now more seriously, you can note that the irrationals ($\mathbb R\setminus\mathbb Q$) are dense, as well all the irrational algebraic numbers ($\sqrt2$ and such). More interestingly the set $\{\sin n\mid n\in\mathbb N\}$ is dense in $[0,1]$ so it can be stretched (or multiplied) into a dense set of $\mathbb R$.

However an important fact is that every countable dense linear order is isomorphic to the rationals, so if your dense set is countable it will not differ too much from the rationals.

Let's construct dense sets in $[0,1]$. We can then get a dense set in $\Bbb R$ by taking the union of dense sets for the intervals $[n, n+1]$ (a dense set for $[n,n+1]$ can be obtained from a dense set of $[0,1]$ by shifting). To make things more interesting, we will find dense sets for which, given any two distinct elements in the set there is a number between them that is not in the set.

For a dense set in $[0,1]$, you can take:

The irrationals in $[0,1]$.

Or: take $[0,1]$. Take its midpoint $1/2$ to be an element in the, to be constructed, dense set. Then take the midpoints of $(0,1/2)$ and $(1/2,1)$ to be elements in the sense set. Then take the midpoints of the four sets obtained by splitting the two prior sets in two...

Or: use any similar, carefully done, construction similar to the preceding example. For instance, you could successively split $[0,1]$ as above (always splitting the previous sets in half), but choose irrationals in each piece.

• It might be noted that the 2nd construction are in fact the dyadic rationals – Tyler Mar 3 '12 at 17:54

Take any irrational number $\alpha$ and consider the set $E = \{n\alpha \bmod 1\ : n \in \mathbb{N}\}$. By the equidistribution theorem this set is uniformly distributed (and thus must be dense) on $[0,1]$. For a set dense on all of $\mathbb{R}$ take $\cup_{n \in \mathbb{Z}} (n + E)$.

Let $a_n$ be a positive sequence going to zero then $p a_n$ with $p\in \mathbb{Z}$ is dense in $\mathbb{R}$.

I guess you have a countable dense set $A$ in mind, since otherwise you could just put $A:={\mathbb R}$. I don't know what your intuition of the real numbers is, but I assume that you are happy with the idea that a real number is an infinite decimal, like $34.5210071856\ldots\$.

The set $$A\ :=\ \bigcup_{r=0}^\infty \left\{{k\over 10^r}\ \bigm| k\in{\mathbb Z}\right\}$$ of finite decimal fractions is a union of countable sets, therefore it is countable. Given any two real numbers $$\alpha:=a_0.a_1\, a_2\, a_3\,\ldots,\qquad\ a_0\in{\mathbb Z},\quad a_k\in\{0,1,\ldots,9\}\ \ (k\geq 1)$$ and $$\beta:=b_0.b_1\, b_2\, b_3\,\ldots,\qquad\ b_0\in{\mathbb Z},\quad b_k\in\{0,1,\ldots,9\}\ \ (k\geq 1)$$ with $\alpha<\beta$ there is a minimal $k\geq0$, call it $k$, such that $a_k<b_k$. Since we have assumed $\alpha<\beta$ the case $$(a_k, a_{k+1}, a_{k+2}\ldots)=(a_k,9,9,9,\ldots)\quad\wedge\quad (b_k , b_{k+1}, b_{k+2}\ldots)=(a_k+1,0,0,0,\ldots)$$ is excluded. There are a few cases to be distinguished, but all in all it is easy to produce a finite decimal expansion $$\xi=x_0.x_1\, x_2\, x_3\,\ldots x_{k-1}\, x_k\ \in A$$ such that $\alpha<\xi<\beta$.

The Liouville numbers are dense in $\mathbb{R}$.

Take every real number with infinite number of occurences of the string '56787773' in the decimal expansion.

• This answer can be improved by changing the number to '8675309' or maybe '42.' I feel like this is a missed opportunity. – Mason Aug 8 '18 at 14:48

Use the axiom of choice to choose one element from each open interval. (I suppose that with this method there's no guarantee you won't get the rationals.)

• You don't need the axiom of choice to choose from every interval. You can constructively choose a rational from the interval. However doing that would result in a subset of the rationals. – Asaf Karagila Mar 3 '12 at 20:05

For any fixed $s>0$,

$$\Bigg\{\sum_{n=1}^\infty\frac{a_n}{n^s}: \{a_n\}_{n=1}^\infty \text{ is a periodic sequence of integers.} \Bigg\}$$ Is a countable set which is dense in the reals. To demonstrate this we can apply checkmath's answer.

• For $s=0$ this seems patently false. Or generally for $s\leq0$. – Asaf Karagila Aug 8 '18 at 15:13
• Analysis is not my game. I don't remember enough to tell you what happens there without spending a positive smidgen of energy, which I currently have none of which to spare for this question. – Asaf Karagila Aug 8 '18 at 15:16