# Supremum of the set $M_x$ for $x\in \mathbb{R}$

Let $x\in \mathbb{R}$ then the set $M_x=\{n\in \mathbb{Z}\;|\; n\leq x\}\neq \varnothing$ is bounded from above, then $M_x$ have a supremum. Donote $[x]=\sup(M_x)$.

On my own, I proved that $[x]\leq x<[x]+1$ for all $x\in\mathbb{R}$.

But what I can not even prove is that $[x]\in\mathbb{Z}, \forall x\in\mathbb{R}$.

Someone can give me some idea of ​​how to do it, using supremum theorems in $\mathbb{R}$?

Thanks.

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$\phi\neq\varnothing$. $\phi$ is a Greek letter and $\varnothing$ is based on a Nordic letter. – Asaf Karagila Nov 6 '12 at 9:25
You probably mean that $M_{x}$ is bounded from above, since it is not a bounded set. – T. Eskin Nov 6 '12 at 9:45
@ThomasE. Exactly – mathsalomon Nov 6 '12 at 9:51
@AsafKaragila Thanks! – mathsalomon Nov 6 '12 at 9:52

Part 1. Intuition. OK, it isn't very hard to just give a formal proof, but I'll try to explain this more intuitively first. If you have a set $M \subset \mathbb{R}$ with a supremum $m = \sup M$, then there are generally two possibilities: either $m \in M$, or $m \not \in M$. In the first case, $m$ is simply the maximum of $M$, and it isn't very interesting. On the other hand, in the second case $M$ doesn't have a maximum, but instead it has a lot of points "right below" $m$. For any positive $\varepsilon$ there should be a point $x$ in $M$ that is below $m$, but above $m-\varepsilon$. So the set $M$ looks very crowded in that area, with lots of points very close to each other.

And this is a clue to your problem. In your case, the set $M_x$ is very sparse, i.e. different points in $M_x$ are far apart from each other, because they are whole numbers. It means that $M_x$ isn't crowded at all, and the second case should be impossible. And in the first case we have $[x] = \sup M_x \in M_x$, which automatically means that $[x] \in \mathbb{Z}$.

Part2. A proof. Now we can build a proof using that intuition. Let $x \in \mathbb{R}$. Since $[x]=\sup M_x$, it follows that there exists an $y \in M_x$ such that $[x]-1 < y \leqslant [x]$. Note that $y$ is a whole number. Suppose that $y \neq [x]$, that is $y < [x]$. Then, by definition of the supremum, there exists an $y' \in M_x$ such that $y < y' \leqslant [x]$. And now we have $y,y' \in M_x$ such that $[x]-1 < y < y' \leqslant [x]$, which means that $|y'-y|<1$. But that is impossible, because $y$ and $y'$ are whole numbers.

This contradiction proves that $y = [x]$, and so $[x] \in M_x \subset \mathbb{Z}$, QED.

PS: yep, not the shortest way to do this. Then again, the definition of $[x]$ in the problem statement using supremums instead of maximums is also kind of strange. But I think that this solution goes well with that definition)

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very good!, I thought that too – mathsalomon Nov 6 '12 at 10:14

Let $x \in \mathbb{R}$, and let $N$ be the greatest element of $M_x$. (The fact that $M_x$ has a greatest element is an easy corollary of the well-ordering principle.) Then $N \geq n$ for all $n \in M_x$, and also $N \in M_x$. It follows that $N$ is the supremum of $M_x$.

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good idea, thanks – mathsalomon Nov 6 '12 at 9:54

Since $[x]<\infty$ is a supremum of $M_{x}$, then (by supremum criteria) for each $n\in\mathbb{N}$ we find $x_{n}\in M_{x}$ so that $[x]-\frac{1}{n}\leq x_{n}$. In particular since $x_{n}\leq [x]$ for all $n\in\mathbb{N}$, then this yields a sequence $(x_{n})_{n=1}^{\infty}\subset M_{x}$ so that $\lim_{n\to\infty} x_{n} =[x]$.

Since $M_{x}\subset \mathbb{Z}$ we now have a Cauchy sequence in $\mathbb{Z}$ so it must be constant after some index, i.e. there exists $k\in\mathbb{N}$ so that $x_{n}$ is constant for all $n\geq k$. Since $\lim_{n\to\infty} x_{n} =[x]$ then this constant must be $[x]$. Since all members of the sequence were members of $\mathbb{Z}$, then in particular $[x]\in \mathbb{Z}$.

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