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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.

Caleb Stanford
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user269334
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1 Answers1

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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, $a_0\in A\cap B$.

Arturo Magidin
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  • 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 08 '12 at 06:17
  • 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 08 '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 08 '12 at 15:58
  • http://math.stackexchange.com/questions/91528/supremum-of-a-set-of-integers is this the proof for "If A is a set of integers and A has a supA, then supA is inside A"? – user269334 Mar 08 '12 at 16:35
  • @user269334: That would be one possibility. Another is to invoke the basic properties of the integers, which include "every nonempty set of integers that is bounded below has a minimum." Take a nonempty set bounded above; multiply every thing by $-1$ and you get a nonempty set bounded below; the minimum of $-A$ is minus the maximum of $A$. – Arturo Magidin Mar 08 '12 at 16:46
  • Another hypothetical question: what if we define \mathbb{Z}_{2} = \left \{ ...,-4,-2,0,2,4,... \right \} A_{2} = \left \{n \in \mathbb{Z}_{2} |n < x \right \} B_{2} = \left \{n \in \mathbb{Z}_{2} |x \leq n+1 \right \}? In that case, even if supA_{2} \in \mathbb{A}_{2},supA \notin \mathbb{B}_{2}?

    Does it have to do with the fact that mathbb{Z} is an inductive set? (so if n is inside Z, n+1 is inside Z?)

    thanks :D

     – user269334 Mar 08 '12 at 22:40
  • let us continue this discussion in chat – user269334 Mar 08 '12 at 22:41
  • @user269334: Please use LaTeX. Using the quote just makes things harder to read. What you write as $\mathbb{Z}2$ should be written $2\mathbb{Z}$, instead. $A{2}$ is still a set of integers, and is bounded above, so it has a maximum. So don't talk about supremum, talk about maximum (it says more). In that case, it is possible for $\max(A)$ to not be in $B$, but it is also possible that it is. For instance, if $x=\frac{1}{2}$, then $\max A = 0$, and $0\in B$. However, if $x=\frac{3}{2}$, then $\max A=0$, but $0\notin B$. Has nothing to do with $\mathbb{Z}$ being inductive. – Arturo Magidin Mar 08 '12 at 22:43
  • @user269334: To use LaTeX, enclose in dollar signs, $2\mathbb{Z}$ to get $2\mathbb{Z}$. It has nothing to do with $\mathbb{Z}$ being inductive, because $2\mathbb{Z}$ is also inductive (under a suitable definition of the successor function). It has to do with how $\mathbb{Z}$ sits inside of $\mathbb{R}$. – Arturo Magidin Mar 08 '12 at 22:45
  • I think I just got it: $\mathbb{A} \subset \mathbb{Z}$ Thus, $maxA \in \mathbb{Z}$ means $(maxA + 1) \in \mathbb{Z}$. Since $(maxA+1) \in \mathbb{Z}$ and (maxA+1) > x, $maxA \in \mathbb{B}$. Is this the right way to do it? – user269334 Mar 09 '12 at 00:04
  • @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 $2n\leq x \lt 2n+2$. – Arturo Magidin Mar 09 '12 at 03:29