Prove $\inf\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots\right\} =0$
Definition I have to satisfy is $\inf X \leq a$, $\forall a \in X$
and if $z \leq a$, $\forall a \in X$ then $\inf X \geq z$
Let the set $X=\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots\right\}$ then since the set $X \not = \emptyset$ it is bounded below by $0$ thus there exists an infimum for the set $X$ which we denote as $y$. It is clear that $y \geq 0$ now if $z \leq a$ $\forall a \in X$ then I need to show for any real number $z>0$, there is an $n$ where $\frac{1}{n}>z$ which will allow me to finish the proof. I know by archidiean property that given any $z>0$ there is an $n \in \mathbb{N}$ such that $\frac{1}{z}<n$ which doesn't give me what I want though.
Any help would be appreciated to help me satisfy the second part.
It is simply seen that $z=0$ is a lowerbound for $X$. Now suppose that $z^\prime > z$ is also a lowerbound, i.e. $\forall a \in X : z^\prime \leq a$. Now, as you suggested, there is an $n \in \mathbb{N}$ such that $0 <\frac{1}{n} < z^\prime$, which contradicts our assumption that $z^\prime$ was also a lowerbound. Hence $z$ is the largest lowerbound of $X$.