Show that exists maximum of $A_x=\{m\in \mathbb{Z}: m\le x\} $

Question

For every $x \in\mathbb{R}$, we define $A_x=\{m\in \mathbb{Z}: m\le x\} $. What i have done thus far is show that $A_x\neq\emptyset$, because if $x\in\mathbb{Z}, A_x=\{x\}$ and as $A_x\subset\mathbb{R}$ and it is bounded above by $x$, so by the Supremum axiom, $A_x$ has a supremum that i called $s$. But i don't know how to show that $s$ is in $A_x$. Any suggestions?

Answer 1

You haven't quite shown that $A_x$ is non-empty in the case that $x$ is not an integer, since in this case $x\notin A_x$. However, for this problem, since the integers are spread apart (so they are discrete), you don't need to use the supremum axiom.

Pick some arbitrary $x\in \mathbb R$.

1. Can you show that there is some integer $n$ such that $x-1 <n\le x$?
2. Can you show that this $n$ is the supremum? (Hint: what can you say about $n+1$?)

Answer 2

I think i prove that $A_x\neq\emptyset$(I use this: "Suppose that $y-x \ge 1$. Prove that there is an integer $k$ such that $x \lt k \lt y$."): "Let $l$ be the largest integer such that $l\le x-\epsilon, \forall\epsilon\ge 1$. Since l is the greatest integer less than or equal to $x-\epsilon$, then $l+1$ must be greater than $x-\epsilon$. So $x-\epsilon \lt l+1$. Since $x-(x-\epsilon) \ge 1$, $x \ge 1+(x-\epsilon)$. Since $l \le x-\epsilon$, then $l+1 \le (x-\epsilon)+1$. Combining these three statements gives:$x-\epsilon \lt l+1 \le (x-\epsilon)+1 \le x$. Therefore $l+1$ is an integer between $x-\epsilon$ and $x$." And it is bounded above by $x$, so it exists supremum, it's correct?