|
Let $A=\left \{ n \in \mathbb{Z} \hspace{1mm}| \hspace{1mm} n \leq x \right \}$. Let $\hspace{2mm} B=\left \{ n \in \mathbb{Z} \hspace{1mm}| \hspace{1mm} x < n+1 \right \}$ Now, how do I know $A\cap B \neq \varnothing$ ? Playing around with "$\mathbb{Z}$ is not bounded" only gave me that $A$ and $B$ exists, but I don't see how I can get $A\cap B \neq \varnothing$ Thanks in advance. |
|||||||
|
|
Note that $A$ is "downward closed" (if $n\in A$ and $m\leq n$, then $m\in A$) and that $B$ is "upward closed" (if $n\in B$ and $n\leq m$, then $m\in B$). Note also that $A$ is nonempty, as is $B$, by the Archimedean property. Also, $A$ is bounded above (by $n+1$ for any $n\in B$), so it has a maximum; and $B$ is bounded below (by $(m-1$ for any $m\in A$), so $B$ has a minimum. Let $a_0$ be the maximum of $A$. Then $a_0\leq x$; we cannot have $a_0+1\leq x$, so $x\lt a_0+1$. Therefore, $a_0\in B$. Thus, $a_0\in A\cap B$. |
|||||||||||
|
|
$B$ is a nonempty set of integers that is bounded below, so it has a least element $n$. Show that $n$ is in $A$. |
||||
|
|
