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To define the function $f(x)=|[x]|$ where $|[x]|$ is the greatest integer that is less or equal to $x$, we need to prove that indeed such an integer exists. In other words,

$$\forall x\in \mathbb{R} \;\;\exists !\, n\in\mathbb{Z}\;\;:\;\;n\leq x<n+1.$$

My first attempt is by defining the set $A=\{y\in\mathbb{Z}| y\leq x\}$. Since $\mathbb{N}$ is not bounded above then $\exists n\in \mathbb{N}(n>-x)$. Then $\exists n'\in \mathbb{Z}(n'<x)$, namely $n'=-n$. Therefore $A\neq\emptyset$. Also $A$ is bounded above by $x$. Then $\sup(A)$ exists.

**Here my problem is how to prove that $\sup(A)\in A$.

Also my question is that I'm not sure if this is enough because $sup(A)$ is unique or do I still need to prove uniqueness.

EDIT:

After all the hints in the answers, chich I appreciate so much, I have arrived to this attempts of solution, please feel free to comment and help me saying if something is wrong:

First approach: considering tree cases ($x<0, x=0, x>0$).

1) If $x=0$ then I can think of $n=0$.

2) If $x>0$ then I can think of the set $B=\{y\in \mathbb{N}|x\leq y\}$. By using the well-ordering principle then $\min(B)-1<x\leq \min(B)$ and then if $x=\min(B) $ I can make $n=\min(B)$; if $\min(B)-1<x<\min(B)$ then $n=\min(B)-1$.

3) if $x<0$ then $-x>0$ and therefore $\exists m \in \mathbb{N}(m\leq -x < m+1)$. Then $\exists n'\in \mathbb{Z}(-n'<x\leq-n'-1)$, namely $n'=m+1$. Again, if $x=-n'-1$ then I can make $n=-n'-1$. If $-n'<x\leq-n'-1$ then $n=-n'$.

Second approach. Since $\mathbb{N}$ is not bounded above then let $x_{0}\in \mathbb{N}$ be such that $x_{0}>\sup(A)$. Let's consider the set $B=\{x_{0}-y: y\in A\}$. Since $B\subseteq \mathbb{N}$ then, by the well ordering principle, $\min(B)$ exists. Moreover $\exists y_{0}\in A\forall y\in A(\min(B)=x_{0}-y_{0}\leq x_{0}-y)$. This means that $\exists y_{0}\in A\forall y\in A (y\leq y_{0})$. Therefore $\max(A)=y_{0}$. In this case I can make $n=\max(A)$.

Caleb Stanford
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Daniela Diaz
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4 Answers4

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Hint: Rather than dealing with $A=\{y\in\mathbb{Z}\mid y<x\}$, try dealing with $B=\{y\in\mathbb{Z}\mid y\leq x\}$.

Why? In general, it is not true that $\sup A\in A$; it is true whenever $x\notin\mathbb{Z}$, but it is false when $x\in\mathbb{Z}$.

Now, think about it this way: let $S:=\sup B$. Since $S$ is the supremum, it must be the case that there is $n\in B$ such that $S-1<n\leq S$.

Can you show that $S\leq x$? If so, you'll have that $n\leq x$. For the other side, note that $S-1<n$ implies $S<n+1$, which implies that $n+1\notin B$; hence $n+1>x$.

Nick Peterson
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  • I make the consideration of $\leq$ insted of $<$ in your hint and also I was trying to follow your sugestions but I couldn't understand what you meant. Anyway, I found two attempts of solutions. Please if you could check and tell me if something is wrong, and also I really like to see your full solution, especially I'd like to know how to justify the existence of $n\in B$ such that $S-1<n\leq S$. Thank you so much. – Daniela Diaz Jul 24 '13 at 15:35
  • This last fact is simply the definition of the supremum: if $S=\sup B$, then $S-\epsilon$ (where $\epsilon>0$) can not be an upper bound on $B$ - because it is smaller than the least upper bound - and so there is some element $n\in B$ such that $n>S-\epsilon$. We know that such an $n$ satisfies $n\leq S$ because $S$ is an upper bound on $B$. – Nick Peterson Jul 24 '13 at 16:56
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First, do you know what real numbers are? In general, you can discuss real numbers without knowing exactly what they are, but I think that when trying to justify some basic properties it is good to be familiar with their definition. There are some equivalent constructions, and if you don't know them I encourage you to check them out.

As for your question, I think that a good approach is to prove the theorem for rational numbers and then deduce it to real numbers since every real number is a limit of rational numbers.

Ido
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$n\le z<n+1$ is not the definition of greatest integer less than or equal to $z$, it's a consequence.

You let $A=\{\,y\in \mathbb Z\mid y\le z\,\}$. You use the Archimedean property (twice) to show $\sup A$ exists. The supremum of a subset of $\mathbb R$ is always unique. To show that $\sup A\in A$ it would be sufficient to show that $A$ is closed, which is the case because $A\subseteq \mathbb Z$.

An alternative method (for brevity here described only for $z>0$) is to note that $\mathbb N\setminus(-\infty,z]$ is nonempty by the Archimedean property, has a smallest element $n$ (by well-order of $\mathbb N$) and that therefore $n-1\le z<n$. From this it follows $n-1$ is the largest integer $\le z$.

Hagen von Eitzen
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  • +1, but someone gave you -1 so you have 0, sorry. Your alternative method took me to my first approach in my edit. Could you please check and tell me if that's what you meant or if that's ok?. – Daniela Diaz Jul 24 '13 at 15:08
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You don't need the $\sup$ for establishing this.

Any construction or description of ${\mathbb R}$ establishes the following facts: $({\mathbb Q},<)\subset({\mathbb R},<)$, and given a real number $\xi$ there are two rational numbers $a$ and $b$ with $a<\xi<b$. Making $a$ a little smaller and $b$ a little bigger we may even assume $a$, $b\in{\mathbb Z}$. The set $\{k\in{\mathbb Z}\>|\>a\leq k\leq b\}$ is finite, and as $\xi<b$ the set $S:=\{k\in{\mathbb Z}\>|\>a\leq k\leq \xi\}$ is finite as well. Therefore $S$ has a maximal element, which we denote by $\lfloor\xi\rfloor$.

Christian Blatter
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