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<x } 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

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