Proving the infimum of $a^n$ for $0 < a < 1$

Question

Started my first real analysis textbook for self-study and was working through the end-of-chapter exercises when I got stuck on this question. From Cummings, Real Analysis, Ch. 1 Exercise 1.27 b):

Fix $a$ $\in$ (0,1). Determine inf($B$) for $B$ = $a^n$: $n$ $\in N$ (i.e. the positive natural numbers).

I know the infimum is 0, and know to prove this I must show it satisfies two criteria - that it is a lower bound, and that given any $\epsilon$ > 0, 0 + $\epsilon$ is not a lower bound of $B$ / $\exists$ $x$ $\in B$ such that $x < 0 + \epsilon$. The first criterion is easy: for non-negative $a$ and $n$, all elements of $B$ must be positive. The second part is what I'm having trouble with: I must show $\exists$ $x$ $\in B$ such that $x < 0 + \epsilon$, as before, which is to say that $a^n <\epsilon$. I'm wondering if I'm allowed to let $a = 0.99 \epsilon$, or something like this, and more generally what a strategy for proving suprema and infima looks like in the non-trivial cases where they are not themselves in the set that they are suprema/infima for.

Answer 1

I would suggest a slightly different approach to showing infimum is $0$. You have shown its bounded below by $0$. So if you prove the sequence $(a^n)_{n \in \mathbb{N}}$ is decreasing for fixed $a$ then we know limit exists from Monotone Convergence Theorem.

So if we call the limit $l$ then we have $$l=\lim_{n \to \infty} a^n=\lim_{n \to \infty} a^{n+1}=a \cdot \lim_{n \to \infty} a^n=a \cdot l$$ And hence $l=0$ (since $a \neq 1$).

Answer 2

If you know the properties of limits quoted by @ScapeProf then that answer works well. If you need a more barebones approach using only facts about inf’s then you need to consider the possibility that $\gamma = \inf(B)$ satisfies $\gamma>0$.

At this point what do you know? You know that there are values $a^n\in B$ very close to $\gamma$. How close to $\gamma$ would $a^n$ need to be before you’d actually have $a^{n+1} < \gamma$?