Show that \(\mathbb{R}\) has the greatest lower bound property.
Show that \(\inf \{1 / n \mid n \in \mathbb{Z}_+\} = 0\) .
Show that given \(a\) with \(0 < a < 1\), \(\inf \{a^n \mid n \in \mathbb{Z}_+\} = 0\) . [Hint: Let \(h = (1 - a) / a\), and show that \((1 + h)^n \geq 1 + nh\) .]
Solution
Let \(A\subset\mathbb R\) be nonempty and bounded below. Then \[ -A=\{-a:a\in A\} \] is nonempty and bounded above. Let \[ s=\sup(-A). \] For every \(a\in A\), \(-a\le s\), so \(-s\le a\); hence \(-s\) is a lower bound of \(A\). If \(\ell\) is any lower bound of \(A\), then \(\ell\le a\) for all \(a\in A\), so \(-a\le-\ell\) for all \(a\in A\). Thus \(-\ell\) is an upper bound of \(-A\), whence \(s\le-\ell\), i.e. \[ \ell\le-s. \] Therefore \(-s=\inf A\), proving the greatest lower bound property.
Every \(1/n\) is positive, so \(0\) is a lower bound. If \(c>0\), the Archimedean property gives \(n\in\mathbb Z_+\) with \[ n>1/c, \] hence \[ 0<1/n<c. \] So no positive number is a lower bound. Therefore \[ \boxed{\inf\{1/n:n\in\mathbb Z_+\}=0.} \]
Let \(0<a<1\) and put \[ h=\frac{1-a}{a}>0, \qquad \frac1a=1+h. \] By induction, \[ (1+h)^n\ge1+nh \] for every \(n\ge1\): the induction step follows from \[ (1+h)^{n+1}\ge(1+nh)(1+h)=1+(n+1)h+nh^2\ge1+(n+1)h. \] Hence \[ a^n=\frac1{(1+h)^n}\le\frac1{1+nh}. \] Given \(\varepsilon>0\), the Archimedean property lets us choose \(n\) with \(1+nh>1/\varepsilon\), so \(a^n<\varepsilon\). Since all \(a^n>0\), it follows that \[ \boxed{\inf\{a^n:n\in\mathbb Z_+\}=0.} \]