Show that $1\in \overline A$

Question

Let $A=\{2^n3^m:m,n\in \Bbb Z\}$. Show that $1\in \overline A$.

In order for $1\in \overline A$ we must find a sequence $x_n$ in $A$ such $x_n\to 1$.

I checked some values of $A$ but could not come up with a sequence .

Can I get some help?

Answer 1

Let $m=n=0$, and we get $2^03^0=1\in A$.

If you require that $mn\neq 0$, we can still construct a sequence: $$ x_n=\frac{3^{\lambda_n}}{2^n}, $$ where $\lambda_n$ is the smallest positive integer such that $\frac{3^{\lambda_n}}{2^n}\geq 1$.

We now prove that $x_n\to 1$. It is easy to check that $$ \lambda_n= \left\lceil\frac{n\ln(2)}{\ln3}\right\rceil. $$

An $n\to \infty$, $\frac{\left\lceil\frac{n\ln(2)}{\ln3}\right\rceil}{\frac{n\ln(2)}{\ln3}}=\frac{\lambda_n}{\frac{n\ln(2)}{\ln3}}\to 1$, so $x_n\to 1$.

You can complete the details of the proof.