# Using an induction argument to show that $\forall n\in\Bbb{N}$, $(n,n+1)\cap\Bbb{N}\neq\emptyset$

An answer using Peano's axioms is provided here: Use an induction argument to prove that for any natural number $n$, the interval $(n,n+1)$ does not contain any natural number.

I'm working through royden's text and it didn't provide those axioms. So I'm wondering if this can be proven without those axiom. The preliminaries given are the algebra of sets, equivalence relations, zorn's lemma, axiom of choice, the field properties, definition of an inductive set, every subset of $\Bbb{N}$ has a least element, archimedean property, and $\Bbb{R}$ is dense. Particularly, I used this definition of an inductive set

A set $E$ of real numbers is inductive provided that it contains $1$ and if $x\in E$, then $x+1\in E$

Proof: (Base case: n=0) Considering $(0,1)$ there cannot be any natural number in the interval since $1$ is the least natural number. (I'm not quite sure here. Royden did mention in a proof that $\Bbb{N}$ is bounded below by $1$ so I thought I could argue this way. Could I?)

(Inductive Case:) Suppose $\forall n\in\Bbb{N}$, $(n, n+1)\cap\Bbb{N}=\emptyset$. Since $\Bbb{N}$ is inductive, $n+1,n+2\in\Bbb{N}$. Also, we note that $\forall q\in(n,n+1)$, then $n<q<n+1 \Leftrightarrow n+1<q+1<n+2$. Keep in mind that $(n,n+1)\cap\Bbb{N}=\emptyset\Rightarrow q\notin\Bbb{N}$. By definition of an inductive set, $q+1\in (n+1,n+2)\Leftrightarrow q\in\Bbb{N}$. Since $q\notin\Bbb{N}$, then $q+1\notin\Bbb{N}$ and $(n+1,n+2)\cap\Bbb{N}=\emptyset$. (I interpreted the definition as an if and only if statement since that's what I was told when dealing with definitions.)

• "Suppose $\forall n\in\Bbb N,(n,n+1)\cap\Bbb N=\emptyset$." Please don't suppose exactly what you want to prove. Commented Jun 9, 2018 at 8:49
• I thought I could use that as an inductive hypothesis? If I cannot, then what should be the proper inductive hypothesis? Commented Jun 9, 2018 at 8:56
• Re your subquestion about th elower bound of $\Bbb N$: Note that $\Bbb N\setminus(-\infty,1)$ is inductive. Commented Jun 9, 2018 at 11:06

## 1 Answer

The following uses these properties of $\Bbb N$:

• Every non-empty subset of $\Bbb N$ has a minimal element. (You listed that among the prerequisites)
• If $x\in\Bbb N$ and $x\ne 1$ then $x-1\in\Bbb N$. (This follows because otherwise $\Bbb N\setminus\{x\}$ would still be inductive)
• $\min\Bbb N=1$. (By the preceding point, $\min\Bbb N\ne 1$ would imply $\min\Bbb N-1\in\Bbb N$)

Let $$S=\{\,m\in\Bbb N\mid \exists n\in\Bbb N\colon n<m<n+1\,\}.$$ The claim we want to prove is equivalent to $S=\emptyset$. So assume $S$ is not empty and let $m=\min S$. Then there exists $n\in\Bbb N$ with $n<m<n+1$. From $m>n\ge 1$, we see that $m-1\in \Bbb N$. Then from $n-1>m-1\ge1$, we see that $n-1\in\Bbb N$. Then $n'<m'<n'+1$ with $n'=n-1$, $m'=m-1$ shows $m'\in S$, contradicting minimality of $m$. $\square$

• I understand that $\Bbb{N}\setminus \{x\}$ should not be inductive. But is the statement "If $x\in\Bbb N$ and $x\ne 1$ then $x-1\in\Bbb N$." simply a restatement that $\Bbb{N}$ is inductive? And that statement follows exactly what statement? thanks Commented Jun 9, 2018 at 16:08