Why $B=\{\frac{1}{n} \mid n \in \mathbb{Z^{+}}\}$ does not have a smallest element?

Question

I am studying from some slides where there's written that $B=\{\frac{1}{n} \mid n \in \mathbb{Z^{+}}\} = \{1,1/2,1/3,1/4, \mid\}$ does not have a smallest element but it does have a g.l.b (infimum)

I understand it has a greatest lower bound, which should be $0$, but I cannot understand why it does not have a smallest element, which in my opinion is incorrect, since $\frac{1}{1}$ or $1$ should be the smallest element.

Answer 1

Let $x \in B$ and suppose by contradiction that $x$ is the smallest element of $B$. Since $x \in B$, there exists $n \in \Bbb Z^+$ such that $x = \frac{1}{n}$. But then $y = \frac{1}{n+1}$ is such that $y <x$ and $y \in B $, a contradiction to the fact that $x$ is the smallest element of $B$.

Answer 2

Suppose $b\in B$ is a smallest element. Then, it must have the property that $b\leq c$ for any $c\in B$. But $b$ must be of the form $b=1/n$ for some positive integer $n$. Then, $c=1/(n+1)\in B$, but $b=1/n>1/(n+1)=c$, a contradiction to $b$ being a smallest element. Therefore, no $b\in B$ can be a smallest element, so $B$ doesn't have any such element.

Note that since $0=\inf\{b\,|\,b\in B\}$, $0$ would be a least element if $0$ were an element of $B$. The problem is that $0\notin B$.

Answer 3

Hint: With the standard ordering of rationals induced by the numerical line $\Bbb{R}$ we have: $$1>\frac{1}{2}>\frac{1}{3}>\cdots>\frac{1}{n}>\frac{1}{n+1}>\cdots >0.$$