Prove that no $m$ exists such that $n < m < n+1$ for positive integers $m$ and $n$.

Question

I am studying basic real analysis using Foundations of Mathematical Analysis by Richard Johnsonbaugh and W. E. Pfaffenberger.

Well-Ordering Theorem:

If $X$ is a nonvoid subset of the positive integers, then $X$ contains a least element; that is, there exists $a \in X$ such that $a \leq x$ for all $x \in X$.

Question:

Use the Well-Ordering Theorem to prove that no $m$ exists such that $n < m < n+1$ for positive integers $m$ and $n$.

Work so far:

I probably want to construct a non-empty set $X$, but I don't know what form $X$ should have. Any hints would be appreciated.

Answer 1

Let $X_n=\{m\in\mathbb{Z},m>0:n<m<n+1\}$ and $Z=\{n\in\mathbb{Z},n>0:X_n\neq\emptyset\}$. You're trying to prove that $Z$ is empty.

Suppose that $Z$ were nonempty. Then, there would be $a\in Z$ such that $a\leq z$ for all $z\in Z$. Consider $X_a$. By definition, $X_a$ would be nonempty so there would exist $z^*$ such that $a<z^*<a+1$. But then $z^*-1<a<z^*$ so that $z^*-1\in Z$ so you would have both $a>z^*-1$ (from $z^*-1<a<z^*$) and $a\leq z^*-1$ (from $z^*-1\in Z$). This is the desired contradiction.

Answer 2

Let $A=\{a\in \mathbb{Z}\;|\;\; a>0\}$.

Prove: The set $B=\{m \in A\;| \exists\;n\in A, \;\;\; n<m<n+1\ \;\; \}$ is empty.

Proof: Fist, we know that 1 is the minimum element of $A$, by definition of the set of positive integers. In particular, by the Well-Ordering Theorem(using your notation), $r=1$.

Let B be non-empty. That is, suppose there exists an integer $m$ such that $n<m<n+1$ for some $n\in A$. By the Well-Ordering Theorem, there exists an element $r_0 \in B$ such that $\forall m \in B$, $r_0\leq m$.

In particular, $n < r_0 < n+1$ for some $n$. Now, add $-n$ to get

$$n+(-n) < r_0+(-n) < n+1+(-n)$$ $$0 < r_0-n < 1$$

Since $n,r_0 \in \mathbb{Z}_{>0}$, and $0<r_0-n$ means $n\not=r_0$ we know $\exists q=r_0-n $ such that $q \in \mathbb{Z}_{>0}$ and $q<1$. By the Well-ordering Theorem, this is impossible as $r=1\le q<1$.

Therefore, no such $m$ exists and $B=\emptyset$.