Dense set in $[0,\infty)$ I want to prove that $M=\{2^a 3^b: a,b \in\mathbb Z\}$ is dense $[0,\infty)$
Therefore I want to show that $\overline{M}=[0,\infty)$
$\overline{M}\subseteq[0,\infty)$ because $M\subseteq[0,\infty)$
For the other direction I have an element $x\in[0,\infty)$, now I need to show that $\exists (x_n)\in\overline{M}:x_n\rightarrow x$, i.e to show $\exists (a_n)(b_n)\in\mathbb Z:x_n=2^{a_n}3^{b_n}\rightarrow x$.
Could you help me with that direction?
EDIT: As discussed in the comments, of course there are different ways proving this result, but I would like to be most interested in the continuation of my approach. If you know different solution approaches it is also interesting to know them and discuss them.
 A: In general if $x,y\in\mathbb N, \ y>1$ and there is a prime $p$ such that $p\mid x$ and $p\nmid y$ then the set $M=\{x^a y^b: a,b \in\mathbb Z\}$ is dense in $[0,\infty)$.
Proof:
Kronecker's theorem can be stated as: If $\xi$ is irrational then the set $\{n\xi+m:n,m\in\mathbb Z\}$ is dense in $\mathbb R$.

Now use the fact that $\frac{\ln x}{\ln y}$ is irrational.
A: Let $x>0$ and let $\varepsilon>0$.  As in P.'s proof, which looks fine to me, we can choose $a,b\in\mathbb Z$ so that 
$$x< 2^a 3^b <x+\varepsilon.$$
To see this, simply take the logarithm and divide through by $\log(3)$ to show that your inequality is equivalent to
$$\frac{\log(x)}{\log(3)} < a\frac{\log(2)}{\log(3)}+b < \frac{\log(x+\varepsilon)}{\log(3)}.$$
Again, as in P.'s answer, the existence of $a$ and $b$ follows from Kronecker's theorem.
Finally, if you really want to construct a sequence of such numbers that converges to $x$, simply choose $a_n$ and $b_n$ so that
$$x< 2^{a_n} 3^{b_n} <x+\frac{1}{n}.$$
Then the sequence defined by $x_n = 2^{a_n} 3^{b_n}$ satisfies your requirement.
