Show given any $x\in\mathbb{R}$ show there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x <n$. I already know that there exists a $n\in\mathbb{N}$ such that $x<n$ by the archimedean property. But to prove that $n-1\leq x$ part I'm not sure. I think I have to create some set such that $n-1$ is a lower bound. But I'm not sure.
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Are you sure this isn't supposed to be 'given any $x\in\mathbb R$'? In any case, the set you should be thinking about is ${a\in\mathbb Z:a\leq x}$. What's the supremum of this set? – Ian Coley Sep 17 '13 at 20:25
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Oh thanks it is the real numbers. The supremum is x. – user60887 Sep 17 '13 at 20:27
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1If it's $x$ then you're fine, but if $x$ isn't an integer, you're going to run into problems. – Ian Coley Sep 17 '13 at 20:27
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There is also the uniqueness of n to be proven as well. – JB King Sep 17 '13 at 20:29
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What if I take your set and switch it where $x>a$ then a lower bound would be a-1? – user60887 Sep 17 '13 at 20:30
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I know but I need to show existence first. – user60887 Sep 17 '13 at 20:30
3 Answers
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Hint: Let $x\ge 0$. By what you wrote, the set $A_x$ of positive integers such that $n\gt x$ is non-empty. Thus $A_x$ has a smallest element. Call that $n$, and show that it does the job.
For negative $x$, use the previous argument to show there is an integer $m$ such that $m-1\le |x|\lt m$. Then $-m\lt x\le -m+1$. If $x\ne -m+1$, we are finished. If $x=-m+1$, make a minor adjustment.
We leave dealing with uniqueness to you.
Remark: We don't need to give special treatment to negative $x$ if we first prove the following small extension of the Least Number Principle: Any non-empty set of integers which is bounded below has a smallest element.
André Nicolas
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You use the Archimedean property twice. Given $x$, the first use gives you an $N$ with $x+1\le N$. Now consider $-x$. The second use of the property gives you an $M$ such that $-x\le M$. This means that $-M\le x< N$. Hence, the set $\{m\in\mathbb Z\mid x< m\}$ is nonempty ($N$ is in it) and is bounded below (by $-M$). So it has a first element. Call it $n$, and note $x<n$. Minimality of $n$ means that $n-1$ is not in the set, so $n-1\le x$. We are done.
Andrés E. Caicedo
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+1 -- I like the point that $x+1 \le N$ is not enough, and that the lower bound $-M \le x$ is necessary. The top answer glosses over that a bit. – Caleb Stanford Jul 06 '16 at 20:37
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Why when you used the Archimedian property for second time you did put less or equal instead of less? – Jonatan Garcia Sep 24 '19 at 07:09
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@6005 , why is $-M\leq x$ necessary? $x$ is by definition already a lower bound of the set ${m\in\mathbb{Z} \mid x<m}$.What am I missing? – Philipp Aug 31 '20 at 12:46
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1@Philipp That is irrelevant: $x$ is also an upper bound for $x$ itself, and yet we need something else to obtain the integer I call $N$. – Andrés E. Caicedo Aug 31 '20 at 14:45
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@Philipp It's exactly as Andrés says: having a real number lower/upper bound doesn't help. The problem requires us to find an integer lower/upper bound. – Caleb Stanford Aug 31 '20 at 15:15
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@6005, ah o.k. I think I got it. Is it legit to envision the idea as follows: we must have an lower bound which is an integer because when we try to find the least element of ${m\in\mathbb{Z}\mid x<m}$ we start "counting" in $+1$-steps from $-m$ onwards and finally arrive at an integer which lies in ${m\in\mathbb{Z}\mid x<m}$- the least element. If we only had a lower bound which is a real number but not an integer then starting to count does not deliver an integer when we arrive at the firt element of ${m\in\mathbb{Z}\mid x<m}$. – Philipp Aug 31 '20 at 19:10
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I'll try to give a proof to the fact that $\forall x \in \mathbb{R} \exists! m \in \mathbb{Z} \left( m - 1 \leq x < m \right)$ using a slightly different approach. First there is the following Lemma.
$\textit{Lemma 1}$. Every nonempty set of integers that is bounded below has a smallest element.
$\textit{Proof of Lemma 1}$. Suppose $S$ is an arbitrary nonempty set of integers that is bounded below. Then using the Axiom of Completeness it is easy to show that $S$ has a greatest lower bound. Let $s = \text{inf} \: S$, and $L = \left\{ x\in \mathbb{R} \mid \forall n \in S \left( x \leq n \right) \right\}$ be the set of all lower bounds for $S$. Clearly $s \in L$. Now consider $s + 1 \in \mathbb{R}$. It is not difficult to show that $s + 1 \notin L$. Then we can find some $n \in S$ such that $s + 1 > n$, so $s \geq n$. Now if $s > n$ then since $s \in L$ we must have $n \notin S$, which is a contradiction. Therefore $s = n \in S$, and since $s \in L$ we can conclude that $s$ is the smallest element of $S$, as required. QED
$\textit{Proof of orignal claim}$. Suppose $x \in \mathbb{R}$. Let $S = \left\{ m \in \mathbb{Z} \mid m > x \right\}$. Clearly $S$ is bounded below. Moreover by the Archimedean Property there is an $n \in \mathbb{Z^+} \subseteq \mathbb{Z}$ such that $n > x$. So $S \neq \emptyset$.
$\textit{Existence}$. By $\textit{Lemma 1}$ we can let some $m \in S$ be the smallest element of $S$. Now suppose $m - 1 > x$. Then since $m - 1 \in \mathbb{Z}$ we get that $m - 1 \in S$. But since $m - 1 < m$ we find a contradiction to the fact that $m$ is the smallest element of $S$. Thus it must be the case that $m - 1 \leq x$, and since $x < m$ we can conclude that $m - 1 \leq x < m$, as required.
$\textit{Uniqueness}$. Suppose $n \in \mathbb{Z}$ and that $n - 1 \leq x < n$. Suppose $n \neq m$. If $n < m$ then since $n \in \mathbb{Z}$ and $n > x$ we get that $n \in S$, and since $n < m$ we find a contradiction to the fact that $m$ is the smallest element of $S$. Now if $n > m$ then $n - 1 \geq m$, so $x \geq m$, which contradicts the fact that $x < m$. Therefore since in either case we found a contradiction we must have that $n = m$, as required. QED
brandao
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