Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any way to declare that "precisely these are the dense subset of $\mathbb{R}$" (well I don't mean the definition of dense set in $\mathbb{R}$) is the set $\{\frac{m}{10^n} : m,n\in \mathbb{Z}, n\ge0\}$ is dense in $\mathbb{R}$?

share|cite|improve this question
No, that seems hopeless. – Apr 20 '12 at 13:52
up vote 5 down vote accepted

$S = \{\frac{m}{10^n} : m,n\in \mathbb{Z}, n\ge0\}$ is dense in $\mathbb{R}$ if $\overline{S} = \mathbb R$.

To see whether this is true pick a point $r$ in $\mathbb R$ and an $\varepsilon$-ball $B(r, \varepsilon)$ around it. The question is whether this ball contains a point of $S$.

Now let's assume $r$ lies somewhere between two integers $a$ and $a+1$. Then $c_0 = \frac{2a+1}{2} = a + \frac{1}{2}$, the point in the middle between $a$ and $a+1$ is in $S$ since $\frac{10a + 5}{10}$ is in $S$. Now if this point is not in $B(r, \varepsilon)$ then either $B(r, \varepsilon)$ lies between $a$ and $c_0$ or $c_0$ and $a+1$. Without loss of generality assume that it lies between $a$ and $c_0$. Then pick the middle point $c_1$ between $a$ and $c_0$ and proceed recursively until you find a $c_k$ that is in $B(r, \varepsilon)$.

So your set is dense in $\mathbb R$.

share|cite|improve this answer

For your first, I don't see how. There are so many of them, most with indescribable structure. To exhibit $2^{\mathfrak c}$ of them, take the complement of the Cantor set. As the Cantor set contains no interval, this is dense. Then add in any subset of the Cantor set, it is still dense.

For your second, if I give you a real number, how can you approximate it better and better with numbers of that form?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.