Let $A = \{ a^n : n \in \mathbb{N} \}$ and assume $0<a<1$. Prove that $\inf A = 0$
We know that $a > 0$ implies $a^n > 0$ and so $0$ is a lower bound of $A$. Suppose $a^n \geq l$ for every $n \in \mathbb{N}$. If we can establish that $0 \geq l$, then $0$ would be the greatest lower bound.
Notice that if $l > 0$ then
$$ \dfrac{1}{a^n} = \left( 1 + \dfrac{1-a}{a} \right)^n \geq 1 + \dfrac{(1-a)n}{a} $$
Note that we can find $n_0$ so that $l \cdot n_0 > \frac{a}{1-a} $ by archimidean principle and so
$$ 1 + \dfrac{(1-a)n_0}{a} > 1 + \dfrac{1}{l} > \dfrac{1}{l} $$
In other words, we have found that $\dfrac{1}{a^{n_0} } > \dfrac{1}{l} $ or that $l > a^{n_0}$ which is a contradiction since $l$ is a lower bound. This forces that $l \leq 0$ and $\boxed{ \inf A = 0 } $ QED
Is this solution correct?
Best Answer
Here's another solution that uses a powerful tool - the monotone convergence theorem. Let $A = \{a^n : n \in \mathbb{N}\}$ where $0 < a < 1$. The sequence $\{a_n\}$ is decreasing and bounded from below, so by the monotone convergence theorem the sequence converges to $\inf A$, i.e. $\lim_{n\to \infty}a^n = \inf A$. Then we have $\inf A = \lim_{n\to\infty} a^{n+1} = a \lim_{n\to\infty} a^n = a \inf A $. But $a< 1$ so $\inf A = 0$. Notice that the use of limit laws is justified since we already know the limit exists.