Every non-empty subset of the integers which is bounded above has a largest element.

I was reading a proof about every non-empty subset of the integers which is bounded above has a largest element, but i have troubles in one step. Here is the proof:

Since $S$ is a non-empty subset of $\mathbb{Z}$(hence $\mathbb{R}$) which is bounded above, by the supremum property $sup S$ exists. Let $w=supS$, then we have to show that $w \in S$, so we suppose $w \notin S$, then we have $w-1<w$, then exists $m\in S$ such that $w-1<m<w$(here is my problem, why can we take $m\in S$ such that happens? we are in $\mathbb{Z}$, so that does not happens). Then he takes $n\in S$ such that $w-1<m<n<w$. The inequality $n<w$ implies that $-w<-n$, and $w-1<m$, if we add up both inequalities we have $-1<m-n$ implies $n-m<1$, then $0<n-m<1$ which is a contradiction because there are no integers between $0$ and $1$ Can anybody explains me if that step is right?? How can I prove this if there is a problem with that step? Thanks!

Solution 1:

I do not think you are able to find both an $m$ and an $n$ such that $m<n$. Once you have the inequality $w-1<m$, then $w<m+1$, showing that $S \subset (-\infty,m+1)$. Since $m\in\mathbb{Z}$, then there is no integer in $(m,m+1)$, so $S\subset(-\infty,m]$. Thus $m\in S$, and $\forall s \in S$, $s \le m$. Thus $m=sup{S}$ and $m \in S$, so $m$ is a maximum of $S$.

Solution 2:

The step you seem to be having problem with: "Let $w=\sup S$, then we have to show that $w\in S$, so we suppose $w\notin S$, then we have $w-1<w$, then there exists $m\in S$ such that $w-1<m<w$. "

This follows from the definition of supremum. Indeed, $w$ is the least upper bound for $S$, and so if there was not any $m\in S$ such that $w-1<m<w$, then it would follow that $m\leq w-1$ for every $m\in S$. But then $w-1$ would an upper bound for the set $S$, which is a contradiction! (Because $w$ was supposed to be the least upper bound for $S$).