User:Abcxyz/Sandbox/Real Numbers/Real Numbers are Dedekind Complete

Theorem
Let $\R$ denote the set of real numbers.

Let $\le$ denote ordering on $\R$.

Then the ordered set $\left({\R, \le}\right)$ is Dedekind complete.

Proof 1
Let $\left({\R, +, \times, \le}\right)$ denote the real numbers, as axiomatically defined as a Dedekind complete totally ordered field.

By definition, $\left({\R, \le}\right)$ is Dedekind complete.

Proof 2
Let $\R$ denote the set of real numbers, as constructed from Cauchy sequences.

Let $\le$ denote the ordering on $\R$.

Let $A \subseteq \R$ be non-empty and bounded above.

Let $a = \left[{\!\left[{\left\langle{a_n}\right\rangle}\right]\!}\right]$ be an upper bound for $A$ in $\R$.

Since a Cauchy sequence is bounded, we have that $\left\langle{a_n}\right\rangle$ is bounded above by some $p_0 \in \Q$.

We have that $\left[{\!\left[{\left\langle{p_0}\right\rangle}\right]\!}\right]$ is an upper bound for $A$ in $\R$.

Let $b = \left[{\!\left[{\left\langle{b_n}\right\rangle}\right]\!}\right] \in A$.

Since a Cauchy sequence is bounded, we have that $\left\langle{b_n}\right\rangle$ is bounded below by some $q_0 \in \Q$.

We have that $p_0 \ge q_0$.

We now define a sequence $\left\langle{u_n}\right\rangle$.

Since $\left({\Q, +, \le}\right)$ is an Archimedean totally ordered group, there exists an $N \in \N$ such that $2^{-n} N > p_0 - q_0$.

We have that $\left[{\!\left[{\left\langle{p_0 - 2^{-n} N}\right\rangle}\right]\!}\right]$ is not an upper bound for $A$ in $\R$.

By the Well-Ordering Principle, the set:
 * $S_n = \left\{{N \in \N: \left[{\!\left[{\left\langle{p_0 - 2^{-n} N}\right\rangle}\right]\!}\right]}\right.$ is not an upper bound for $A$ in $\left.{\R}\right\}$

has a smallest element $\sigma_n \ne 0$.

We define the sequence $\left\langle{u_n}\right\rangle$ as:
 * $u_n = p_0 - 2^{-n} \left({\sigma_n - 1}\right)$

We claim that $u = \left[{\!\left[{\left\langle{u_n}\right\rangle}\right]\!}\right]$ is the supremum of $A$ in $\R$.

First, we show that $\left\langle{u_n}\right\rangle$ is a Cauchy sequence.

Since $2 \sigma_n \in S_{n+1}$, we have that $\sigma_{n+1} \le 2 \sigma_n$.

Hence:
 * $u_n - u_{n+1} = 2^{-n-1} - 2^{-n-1} \left({2 \sigma_n - \sigma_{n+1}}\right) \le 2^{-n-1}$

Since $0 \le 2 \sigma_n - 2 \notin S_{n+1}$, we have that $\sigma_{n+1} > 2 \sigma_n - 2$.

Therefore, $\sigma_{n+1} \ge 2 \sigma_n - 1$.

Hence:
 * $u_n - u_{n+1} = 2^{-n-1} \left({\sigma_{n+1} - 2 \sigma_n + 1}\right) \ge 0$

Let $\epsilon \in \Q_{>0}$.

By Power of Number Less Than One, there exists an $N \in \N$ such that:
 * $\forall n \in \N: n > N \implies 2^{-n} < \epsilon$

Let $m, n \in \N$, $N < m < n$.

Then:

We show that $u$ is an upper bound for $A$ in $\R$.

Suppose $x = \left[{\!\left[{\left\langle{x_n}\right\rangle}\right]\!}\right] \in A$.

Let $\epsilon \in \Q_{>0}$.

Since $\left\langle{u_n}\right\rangle$ is a Cauchy sequence, there exist $M, N \in \N$ such that:
 * $\forall m, n \in \N: m, n > M \implies \left\vert{u_m - u_n}\right\vert < \frac 1 2 \epsilon$
 * $\forall n \in \N: n > N \implies x_n < u_{M+1} + \frac 1 2 \epsilon$

Let $N' = \max {\left\{{M, N}\right\}} \in \N$.

It follows that:
 * $\forall n \in \N: n > N' \implies x_n < u_{M+1} + \frac 1 2 \epsilon \le u_n + \epsilon$

Hence, $x \le u$.

Let $v = \left[{\!\left[{\left\langle{v_n}\right\rangle}\right]\!}\right]$ be an upper bound for $A$ in $\R$.

We show that $u \le v$.

Let $\epsilon \in \Q_{>0}$.

Since $\left\langle{v_n}\right\rangle$ is a Cauchy sequence, there exists an $M \in \N$ such that:
 * $\forall m, n \in \N: m, n > M \implies \left\vert{v_m - v_n}\right\vert < \frac 1 3 \epsilon$

We have that $\left[{\!\left[{\left\langle{v_n}\right\rangle}\right]\!}\right] \le \left[{\!\! \left[{\left\langle{v_{M+1} + \frac 1 3 \epsilon}\right\rangle}\right]\!\!}\right]$, so $\left[{\!\!\left[{\left\langle{v_{M+1} + \frac 1 3 \epsilon}\right\rangle}\right]\!\!}\right]$ is an upper bound for $A$ in $\R$.

By Power of Number Less Than One, there exists an $m \in \N$ such that $2^{-m} < \frac 1 3 \epsilon$.

We have that $p_0 - 2^{-m} \sigma_m < v_{M+1} + \frac 1 3 \epsilon$.

Therefore, $u_m < v_{M+1} + \frac 2 3 \epsilon$.

Let $N = \max {\left\{{M, m}\right\}} \in \N$.

Let $n \in \N$, $n > N$. Then:
 * $u_n \le u_m < v_{M+1} + \frac 2 3 \epsilon < v_n + \epsilon$

Hence, $u \le v$.

Proof 3
Let $\R$ denote the set of real numbers, as constructed from Dedekind cuts.

Let $A \subseteq \R$ be non-empty and bounded above.

Let $\displaystyle \beta = \bigcup A$ be the union of $A$.

We claim that $\beta$ is the supremum of $A$ in $\R$.

First, we show that $\beta$ is the supremum of $A$ in the power set $\mathcal P \left({\Q}\right)$.

From Subset of Union, we have that $\beta$ is an upper bound for $A$ in $\mathcal P \left({\Q}\right)$.

Suppose that $u$ is an upper bound for $A$ in $\mathcal P \left({\Q}\right)$.

Then $\forall \alpha \in A: \alpha \subseteq u$.

From Union Smallest, we have that $\beta \subseteq u$.

Now, we show that $\beta \in \R$.

Since $A$ is non-empty and all of its elements are non-empty, it follows that $\beta$ is non-empty.

Let $v$ be an upper bound for $A$ in $\R$.

Then $\beta \subseteq v$.

Since $v \ne \Q$, we have that $\beta \ne \Q$.

Suppose $q \in \beta$.

Then we can choose $\alpha \in A$ such that $q \in \alpha$, and we can choose $r > q$ such that $r \in \alpha \subseteq \beta$.

If $p \in \Q$ and $p < q$, then $p \in \alpha \subseteq \beta$.

Proof 4
Let $\left({\R, \le}\right)$ denote the ordered set of real numbers, as defined as the Dedekind completion of the rational numbers.

By definition, $\left({\R, \le}\right)$ is Dedekind complete.