Showing there is no natural number between two consecutive natural numbers

Question

I want to show that:

$x\subset S(x)$ where $S$ is the Successor function
and
$\not\exists z:x\subset z\subset S(x)$

These are obvious results, but the relation of $m<n\iff m\in n$ is given as a definition, this definition could be directly applied but this question sort of proves it!

The $x\subset S(x)$ is easy enough! The other is more difficult.

By the way the definition of $S(x)=x\cup \{x\}$

My reason for not writing what I have so far is that it's tonnes of $\{$ and $\}$

Answer 1

$\def\s#1{\{#1\}}$ Lemma. If $x\subset z\subset x\cup \s y$, then either $z=x$ or $z=x\cup\s y$.

1. Suppose $y \in z$. Then $\s y\subset z$, and since $x\subset z$, then $x\cup\s y\subset z$ and we are done.
2. Or suppose $y\notin z$. Then let $t\in z$. Since $z\subset x \cup \s y$ then either $t\in x$ or $t\in \s y$. But $y\notin z$ so $t\ne y$. So $t\in x$, so $z\subset x $ and we are done.

We want to show that if $x\subset z \subset S(x)$, then either $z=x$ or $z = S(x)$. Take $y=x$ in the lemma.

Answer 2

If x$\subset$ z$\subset$ x $\cup${x} then z=z$\cap$(x$\cup${x}) as z$\subset$ x$\cup${x} $\implies$ z=(z$\cap$x)$\cup$(z$\cap${x})=x$\cup$(z$\cap${x}) as x$\subset$z now z$\cap${x}={x} or $\emptyset$ then z$\cap${x}={x} $\implies$ z=S(x) and z$\cap${x}=$\emptyset$ $\implies$ z=x. Hence there exist no set that is strictly between x and S(x)

Answer 3

Suppose that $n\in\mathbb{N}$. Now, suppose for contradiction that there is a $k\in\mathbb{N}$ such that $n<k<s(n)$. Since $s(n)=n\cup\{n\}$, this means that $n<k<n\cup\{n\}$.

Since $k<n\cup\{n\}$, we must have $k\in n\cup\{n\}$ by definition. Since $k\in n\cup\{n\}$, $k\in n$ or $k\in\{n\}$. We consider each case separately.

If $k\in n$, then $k<n$ by definition. However, this contradicts the fact that $<$ is asymmetric on $\mathbb{N}$. Thus, we cannot have $k\in n$.

If $k\in\{n\}$, then $k=n$. However, since $n<k$ and $k=n$, we have $n<n$, which contradicts the fact that $<$ is irreflexive on $\mathbb{N}$. Therefore, we cannot have $k\in\{n\}$ either.

Because of these contradictions, we reject our initial assumption and conclude that there is no $k\in\mathbb{N}$ such that $n<k<s(n)$.$\Box$