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I have another question about proving "For every real number $x$, there's exactly one integer $n$ such that $n \leq x\lt n+1$".

Let $A=\left \{ n \in \mathbb{Z} \mid n \leq x \right \}$. Let $\hspace{2mm} B=\left \{ n \in \mathbb{Z} \mid x < n+1 \right \}$

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.

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    [Related](http://math.stackexchange.com/questions/117734/proof-of-greatest-integer-theorem-floor-function?rq=1)2016-07-17

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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, $a_0\in A\cap B$.

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    @user269334: Yes; if you were to use your $2\mathbb{Z}$ instead, you have the same argument works if you are looking for the unique $n\in 2\mathbb{Z}$ such that $2n\leq x \lt 2n+2$.2012-03-09