What is the lower bound of the subset $2^n,\; n\in\mathbb{N}$

Question

Let:

$$ A = \{2^n,\; n\in\mathbb{N}\},\quad A\subset \mathbb{R} $$

Is the lower bound:

• $(-\infty,0]$
• $(-\infty,1]$
• $(-\infty,1)$

?

I think it can be the first because $\min_{A}=1,\;\inf_{A}=1$, according to the definition of lower bounds.

Answer 1

Let $a$ denote the sequence indexed by $\mathbb{N} = \{0,1,2,\cdots\}$, with defining property $$a_n = 2^n,$$

and define $A = \{a_n \mid n \in \mathbb{N}\}.$

If I'm not mistaken, the problem is to find the set of all lower bounds of $A$. Observe that the bigger $n$ gets, the bigger $a_n$ gets. Thus $a_0$ ($=1$) is the least element of $A$. This implies that $(-\infty,1]$ is the set of all lower bounds of $A$.