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Let $0 < x$. Show that there is a unique $m \in \mathbb{N}$ such that $m-1 \leq x < m$. Hint: Consider the set $\{ n \in \mathbb{N} : x < n\}$ and use the well-ordering of $\mathbb{N}$.

The textbook definition says $\mathbb{N}$ is well ordered and every nonempty subset of $\mathbb{N}$ contains a least element. For example, if $A$ is a nonempty subset of $\mathbb{N}$, then there is an $a_0$ in $A$ such that $a_0 \leq a$ for all $a \in A$.

My professor gave a further hint that from $m-1 \leq x < m$, we should also consider another variable $m'$, and say $m'-1 \leq x < m'$. My professor is now asking us to show that $m'=m$, to show that $m$ is unique.

Despite all this information, I cannot seem to piece it together and show the uniqueness of $m$. Which proof technique I should use that works best here?

E W H Lee
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user176071
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3 Answers3

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Hints: Let $A = \lbrace n \in \mathbb N:x<n\rbrace$. $A$ is clearly non-empty so by the well-ordering principle, $A$ has a least element $a \in A$.

  1. Since $a \in A$, $x<a$
  2. What about $a-1$? Is $x<a-1$ or $x \ge a-1$? Why?

For the uniqueness, if $m-1\le x \lt m$ and $m'-1 \le x \lt m'$ , then without loss of generality, we can assume that $m \le m'$. Then $$m-1\le m'-1\le x \lt m \le m'$$Can this happen? If so how?

Mathmo123
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This is an existence and uniqueness proof. First we show existence. Let $0<x$ and let $N_{x}=\{n \in \mathbb{N}:x<n\}$ We know $N_{x}$ is non-empty because $\lceil x \rceil \in \mathbb{N}$ and $x \leq \lceil x \rceil<\lceil x \rceil+1$ so $\lceil x \rceil+1 \in N_{x}.$ By the well ordering principle, we know $N_{x}$ contains a least element because $N_{x} \subseteq\mathbb{N}$. Let $m \in N_{x}$ be the least element. Since $m \in N_{x}$ we know $x<m$. Now for the sake of contradiction suppose $x<m-1$. Then $m-1 \in N_{x}$ by definition of $N_{x}$. This is a contradiction, because $m-1<m$ yet we already have $m$ as the least element of $N_{x}$

For uniqueness let's do contradiction again. Suppose there is some other $m' \in N_{x}$ such that $m'-1\leq x <m'$. We already have $m$ as the least element of $N_{x}$ so $m'>m$. Further, as we are dealing with natural numbers we know $m+1 \leq m'$, as $m'$ must be at least one integer value larger than $m$. Clearly $x<m+1$ because $x<m$. We also have $m'-1<x$ and $m \leq m'-1$ is obtained from $m+1 \leq m'$. It follows that $m<x$. This is a contradiction because we already know $m>x$. Thus, no such $m'$ can exist. $m$ is the unique element that works for our $x>0$.

graydad
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Take the set $A=\{n\in\Bbb N:n\leq x\}$. This set is non-empty ($0$ belongs to it) and bounded above ($x$ is an upper bound). So it follows that $A$ has a least upper bound. Let's call it $\alpha$. Clearly $\alpha$ is an element of $\Bbb N$. [Because if $\alpha$ does not belong to $\Bbb N$, then (by definition of least upper bound) it must be an accumulation point for $\Bbb N$. But there are no accumulation points for $\Bbb N$ other than the elements of $\Bbb N$ itself.] Let $m=\alpha + 1$. We claim that $m-1\leq x<m$. In fact $m-1=\alpha\leq x$ because x is an upper bound for the set $A$; and $m=\alpha + 1> x$ for if it wasn't then $\alpha$ wouldn't be an upper bound of $A$.

It remains to show that this $m$ is unique. So suppose there exists another natural number $m'$ such that $m'-1\leq x<m'$. Without loss of generality let's say that $m\leq m'$. Then we have that $m'-1<m\leq m'$. But between any two consecutive natural number there isn't any other natural number. Therefore $m'=m$.

User
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