# Completeness of the Real numbers

Cut: A cut in ${ \mathfrak{Q} }$ (The set of rational numbers) is a set ${ \alpha \subset \mathfrak{Q} }$ such that

${ \alpha \neq \Phi }$.
${ \alpha^{c} \neq \Phi }$.
If ${ r \in \alpha }$, ${ s \in \mathfrak{Q} }$ and ${ r \leq s }$, then ${ s \in \alpha }$.
${ \alpha }$ does not have a least element.

Real numbers: The set of real numbers ${ \mathfrak{R} }$ is defined as the set of all cuts in ${ \mathfrak{Q} }$.

If ${ \alpha }$, ${ \beta }$ are cuts, then we say that ${ \alpha \leq \beta }$ iff ${ \beta \subset \alpha }$.

Theorem 1 ${ \mathfrak{R} }$ is complete.

Proof: Let ${ B \subset \mathfrak{R} }$ be bounded below.

Now consider ${ \beta_{1}=\bigcup_{\beta \in B} \beta }$.

We first show that ${ \beta_{1} }$ is a cut.

Since ${ \beta \neq \Phi }$ as it is a cut and there exists atleast one ${ \beta \in B }$, ${ \beta_{1} \neq \Phi }$.

${ \forall r \in \beta_{1} }$, ${ \exists \hat{\beta} \in B }$ such that ${ r \in \hat{\beta} }$.

${ \forall s\in \mathfrak{Q} }$ and ${ s>r }$, ${ s \in \hat{\beta} }$ and hence, ${ s \in \beta_{1} }$.

If ${ \beta_{1} }$ had a least element, say x, then since ${ x \in \beta_{1} }$, ${ x \in \beta }$ for some ${ \beta \in B }$. But for every ${ x \in \beta }$, ${ \exists y \in \beta }$ and hence ${ \exists y \in \beta_{1} }$ such that ${ y since ${ \beta }$ does not have a least element. Hence, ${ \beta_{1} }$ does not have a least element. Therefore, ${ \beta_{1} }$ is a cut.

Now, for every ${ \beta \in B }$, ${ \beta \subset \bigcup_{\hat{\beta}\in B} \hat{\beta} }$. Hence, ${ \beta \subset \beta_{1} }$. Hence, ${ \beta_{1} \leq \beta }$ ${ \forall \beta \in B }$. Therefore, ${ \beta_{1} }$ is a lower bound for B.

Consider any other lower bound, say ${ \gamma }$ for B. Now, ${ \gamma \leq \beta }$ and hence ${ \beta \subset \gamma }$ ${ \forall \beta \in B }$.

Hence, ${ \bigcup_{\beta \in B} \beta \: \subset \gamma }$. This implies that ${ \beta_{1} \subset \gamma }$ and ${ \beta_{1} \leq \gamma }$. Hence, ${ \beta_{1} }$ is the greatest lower bound for B.

We have shown that for any arbitrary set ${ B \in \mathfrak{R} }$, there exists a greatest lower bound. Hence, for every ${ B \in \mathfrak{R} }$, there exists a greatest lower bound. Hence, ${ \mathfrak{R} }$ satisfies the greatest lower bound property. Therefore, ${ \mathfrak{R} }$ satisfies the least upper bound property. This implies that ${ \mathfrak{R} }$ is complete.

$\Box$