# Another question on proving “For every real number $x$, there's exactly one integer $n$ such that $n \leq x\lt n+1$”

Let $A=\left \{ n \in \mathbb{Z} \hspace{1mm}| \hspace{1mm} n \leq x \right \}$. 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 $A\cap B \neq \varnothing$

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The answer to this really depends on what you take to be the foundational assumptions. Do you have in mind a certain axiomatization of the reals and integers? A certain construction of the reals such as Dedekind cuts? – Ben Crowell Mar 8 '12 at 4:27
I'm doing Apostol Calculus book right now...I haven't yet seen Dedekind, I think – user269334 Mar 8 '12 at 4:39

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, $a_0\in A\cap B$.
There's stuff you don't need; we didn't really use the downward-closed or upward-closed property, but it should tell you that what you have is $A\cup B=\mathbb{Z}$; and by symmetry, you can consider the minimum of $B$ instead of the maximum of $A$. – Arturo Magidin Mar 8 '12 at 6:17
A quick question here: how do we know a_{0} \in A ? I mean, we can't assume supA \in A...or does the answer you gave me somewhat implies this? – user269334 Mar 8 '12 at 12:21
@user269334: $a_0$ is the maximum of $A$; maxima are always in the set. $A$ is a set of integers, not reals; if it has a supremum, then it has a maximum. – Arturo Magidin Mar 8 '12 at 15:58