Infimum of a Simple Set

Question

I am struggling in proving infimum of a simple set like $S=\left\{\left(\frac{2}{3}\right)^{n}\;;\;n\in\mathbb{N}\right\}$ what I would do first is notice that $n\geq 1$ would imply that $2^{n}\geq 2\implies \left(\frac{2}{3}\right)^{n}\geq \frac{2}{3^{n}}$ and so one can see that $0$ is the infimum but I don't know how to properly use the Archimedean property here to show that $0$ is the greatest lower bound of $S$.

Answer 1

The set $S$ consists only of positive numbers, such that no positive number is an $\inf$ (i.e., for any positive $\varepsilon$ number there is a element of $S$ that is smaller than $\varepsilon$). To show that one has to find for any $\varepsilon$ that an $n$ exists such that $$ \left(\frac{2}{3}\right)^{n} \le \varepsilon \\ \overset{\ln(\cdot)}{\implies}\quad n\ln\left(\frac{2}{3}\right) \le \ln(\varepsilon) \quad\overset{(\cdot)/\ln\left(\frac{2}{3}\right)}{\implies}\quad n \ge \frac{\ln(\varepsilon)}{\ln\left(\frac{2}{3}\right)} $$ the last inequality sign was reversed because of division by $\ln(2/3)$, which is negative.

That means that the $\inf$ in non-positive (negative or zero.) The inf can't be negative because for any negative there is greater negative as a lower bound. So it remains to be zero