Find $\sup$ of an infinite set

Question

Let $A = \{ 0.2, 0.22, 0.222, \dots\}$.

I've noticed that $n$-th element of the set is equal to $2\cdot\left({10^{-1}}+ ... +{10^{-n}}\right)$ which is less than $2\cdot\left(\frac{1}{1- \frac{1}{10}}\right) = \frac{2}{9}$.

Now I want to show that $\sup A = \frac{2}{9}$ and I really got stuck at this point. I know from the definition of supremum: $\forall \varepsilon\ \exists x\ x > \sup A - \varepsilon$.

And it seems really obvious that for any $\varepsilon$ there is always an opportunity to choose such $x$, but I don't know how to write this fact formally.

Answer 1

Let $a_n=2\cdot\left({10^{-1}}+ ... +{10^{-n}}\right)$ then note that $$\frac{2}{9}-a_n=2\sum_{k=n+1}^{\infty}{10^{-k}}=\frac{2}{10^n}\sum_{k=1}^{\infty}{10^{-k}}=\frac{2}{9\cdot10^n}<\varepsilon.$$ as soon as $n>\log_{10}\left(\frac{2}{9\varepsilon}\right)$.

Answer 2

You could literally solve for $x$ given an $\epsilon$. If $\epsilon = 10^{-n}$ then $x=n$ appears to work. You could then always round $\epsilon$ to the greatest power of 10 less than $\epsilon$. (ie $10^{\lfloor \log_{10} \epsilon \rfloor}$)