Exists Rational Cut Between two Cuts

Theorem

Let $\alpha$ and $\beta$ be cuts.

Let $\alpha < \beta$, where $<$ denotes the strict ordering on cuts.

Then there exists a rational cut $r^*$ associated with the rational number $r$ such that:

$\alpha < r^* < \beta$

Proof

Let $\alpha < \beta$.

Then there exists a rational number $p$ such that $p \in \beta$ and $p \notin \alpha$.

Let $r \in \Q$ such that $r > p$ and $r \in \beta$.

Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$.

Because $p \in r^*$ and $p \notin \alpha$, we have that $\alpha < r^*$.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.29$. Theorem