Find the supreme and infimum of the following set, $A=\left \{\dfrac{1}{2^n}: n\in \Bbb N\right \}$. I know that the upper bound is $\frac{1}{2}$ and the lower bound is $0$, since the set is:
$$\left \{\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},...,\dfrac{1}{2^n}\right \}$$
and the limit of $\dfrac{1}{2^n}$ tends to zero.
To find the  $sup(A)$ and $inf(A)$. I know I should reverse engineer the result using $\varepsilon > 0$ and the archimedes principle. But I am having a really hard time doing this because $n$ is in the exponent.
How do I proof that those are in fact the supreme and infimum of this set?
 A: Since $A$ is a non-empty and bounded set, both supremum and infimum of $A$ exists in $\mathbb{R}$. Notice that $\frac{1}{2^{n+1}} < \frac{1}{2^n}$ (so the terms of $A$ decrease with $n$) and from this $\max A = 1/2$. Therefore: $\sup A = 1/2$.
Related to the infimum, let's prove it's indeed $0$. Let $\varepsilon >0$, then we look for $n_0 \in \mathbb{N}$ such that $2^{-n} < \varepsilon$. Taking logarithms we get: $n \ln (1/2) < \ln (\varepsilon)$. Finally, we have $\ln (1/2) = -\ln(2) <0$, so the inequality can be written as: $n> - \ln(\varepsilon) / \ln(2)$ (Notice that this inequality makes total sense since $\ln \varepsilon$ is negative for small $\varepsilon$). The Archimedes Principle applies for finding $\color{red}{n_0} = \lfloor -\ln(\varepsilon)/\ln(2) \rfloor + 1\in \mathbb{N}$ (as a direct consequence of $\mathbb{N}$ being not bounded above) such that $1/2^n < \varepsilon$. Finally, by the classical result:
$$I=\inf A \Longleftrightarrow \forall \varepsilon >0, \exists a\in A: I+\varepsilon < a$$ , we have $\inf A = 0$ (for arbitrary $\varepsilon >0$, take $a=1/2^{\color{red}{n_0}}$). I hope this helps!
A: Note that, for all $n\in\mathbb N_{\ge 2}$ , we have
$$\frac 12-\frac 1{2^n}=\frac {2^{n-1}-1}{2^n}>0.$$
Because, $2^{n-1}>2^{0},\,\forall n\in\mathbb N_{\ge 2}$. This implies,
$$\sup\left \{\frac{1}{2^n}: n\in \mathbb N\right \}=\frac 12$$
Obviously, infimum should be nonnegative and cannot be greater than $\frac 12$. Suppose that, $0\leq A\leq \frac 12$ is an infimum. This implies $A\leq\frac 1{2^n}$ for all $n\in\mathbb N$.
But, on the other hand
$$\begin{aligned}A>\frac 1{2^n}\, \wedge \, n\in\mathbb N&\implies 2^nA>1\\
&\implies 2^n >\frac 1A\\
&\implies n\ln 2>-\ln A\\
&\implies n>-\frac {\ln A}{\ln 2}\\
&\implies n\ge \Big{\lfloor}-\frac {\ln A}{\ln 2}\Big{\rfloor}+1\end{aligned}$$
This leads, if $n\ge \Big{\lfloor}-\frac {\ln A}{\ln 2}\Big{\rfloor}+1$,
then $A>\frac 1{2^n}$. A contradicton.
This means,
$$\inf\left \{\frac{1}{2^n}: n\in \mathbb N\right \}=0.$$
