The proposition is: If x is an arbitrary real number, prove that there is exactly one integer n which satisfies the inequalities n ≤ x < n + 1. This n is called the greatest integer in x and is denoted by [x]. For example, [5] = 5, [5/2] = 2, and [−8/3] = −3.
These are the proof process:
My question is why in the end, the integer n + 1 not ∈ S implies that x < n + 1 ? x is an upper bound and n+1 is an upper bound. How do I know that n+1>x from the implictaion?
Rewriting it to maybe illuminate things:
We define $S = \{m \in \mathbb{Z} : m \leq x\}$. Note that $x$ is an upper bound of $S$ by construction; take $y \in S$, then $y \leq x$ automatically. Note that $S$ is nonempty (apparently an exercise). Consider $\sup S = n$. This is the least upper bound. So it is an upper bound (meaning that for all $y \in S$, $y \leq n$) and if $z$ is any other upper bound, $n \leq z$ (it is the smallest upper bound). The claim is that $n$ is an integer. By a property of supremums, if we take $\epsilon = 1$, there is some $m \in S$ so that $n-1 < m \leq n$. Now if there is some $k \in S$ so that $m < k$, then $k -m \in \mathbb{Z}$ and $k -m > 0$ strictly. The strictness and the fact it's an integer means $k-m \geq 1$. Now $k \geq m+1 > (n-1)+1 = n$, a contradiction since $n$ is supposed to be an upper bound so $n < k \leq n$. Thus $k \leq m$ for all $k \in S$, and so by the least upper bound property we have that $n \leq m \leq n$. So $m = n \in \mathbb{Z}$.
Now $n$ is an upper bound, so $n + 1 \notin S$ (for if it were, then $n+1 \leq n$, which is a contradiction). Since $n \in \mathbb{Z}$, we have $n+1 \in \mathbb{Z}$. Write $S^c = \{m \in \mathbb{Z} : m > x\}$. If $n+1 \notin S$, this means that $n+1 \in S^c$, so $n+1 > x$. This gives us $n \leq x < n+1$.