# Find countable set that is not in the additive, multiplicative closure of a finite set.

I would like to find a sequence $a_0, a_1, a_2, \dots$ of elements of $[0,1]$, such that there is no finite set $B = \{b_0, \dots , b_n\}$ of elements of $[0,1]$ such that every $a_n$ can be written as a finite combination (using multiplication and summation) of elements of $B$.

For example, I think that $a_n = \frac{1}{n}$ should work. At least it is not possible to find a finite set of rationals from which you can make every number $\frac{1}{n}$ just by multiplication and addition. However, it becomes a bit more difficult if you also consider irrationals.

Although, I think it should be possible to solve it in an easier way, I also tried something with field extensions (and the degree). However, here it becomes difficult if $B$ consists of transcendental numbers.

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We know that $\mathrm{tr\ deg}(\mathbb R/\mathbb Q)$ is infinite. Let $A=\{a_n: n\geq1\}\subset\mathbb R$ be an infinite algebraically independent subset over $\mathbb Q$. Changing $a_n$ by $a_n-\lfloor a_n\rfloor$ we can suppose that each $a_n\in[0,1)$. For any subset $B\subseteq [0,1]$, the ring closure $\mathbb Z[B]$ of $B$ is contained in $\mathbb Q(B)$, the subfield of $\mathbb R$ generated by $B$ over $\mathbb Q$, so if $\mathbb Z[B]$ contains $A$, then $\mathbb Q(B)$ also contains $A$. Consequently, by well-known properties of the transcendence degree we have the inequalities $|B|\geq\mathrm{tr\ deg}\bigl(\mathbb Q(B)/\mathbb Q\bigr)\geq |A|$, which shows that $B$ must be infinite.
Thanks! This works. However it still feels a bit cumbersome to use transcendental extensions and their degree to prove this statement. Does someone know if the example with $a_n= \frac 1 n$ works? – Tessa Feb 4 '13 at 7:59
Assuming you only multiply by naturals, any set that has $0$ as a limit point will work. It will have an element smaller than any element of $B$.