TFAE: Completeness Axiom and Monotone Convergence Theorem

I'm looking to write a proof that shows

Completeness Axiom / L.U.B. Property of $\mathbb{R}$$\iff$ Monotone Convergence Theorem.

However, it's quite perplexing to me how a proposition about real numbers can imply such an in-depth result about sequences and series.

Could anyone offer me some starting points? I'm thinking by contradiction for $\Leftarrow$, but I have no idea how to do $\Rightarrow$.

Solution 1:

"$\Leftarrow$": Short version: Below is a proof that MCT implies the LUB property. The basic idea is to take the set $U$ of upper bounds of a nonempty set $A$ that is bounded above, and to construct an increasing sequence $(x_k)$ with no term in $U$, such that $x_k+\varepsilon_k$ is in $U$ for all $k$, and $\varepsilon_k\to 0$. MCT implies that $(x_k)$ converges, and the limit can be shown to be $\sup A$. If you are only looking for hints, read no further.

• Lemma: MCT $\Rightarrow$ Archimedean Property.
  Proof by contradiction: If $\mathbb{N}$ is bounded, then the sequence $(n)_n=(1,2,3,4,\ldots)$ converges by hypothesis to some real number $x$, and then $(n+1)_n$ converges to $x$ and to $x+1$, a contradiction.$\square$

Let $A$ be a nonempty set of real numbers that is bounded above, and let $U$ be the set of upper bounds of $A$.

• Claim 1: For all $\varepsilon>0$, $U-\varepsilon:=\{u-\varepsilon:u\in U\}$ is not contained in $U$.
  Proof by contradiction: Let $\varepsilon>0$. If $U-\varepsilon\subseteq U$, then $U-n\varepsilon\subseteq U$ for all $n$. There exists $u\in U$, and $U$ therefore contains the ray $\{x:x>u\}$. This implies by the Archimedean property that $\mathbb R=\bigcup\limits_{n=1}^\infty (U-n\varepsilon)\subseteq U$, which contradicts $A\neq\emptyset$.$\square$

• Claim 2: $\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)\subseteq U$.
  Proof: Suppose that $x$ is in $\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)$, and that $y>x$. By the Archimedean property there exists $n\in\mathbb N$ such that $x+\frac{1}{n}<y$. But $x+\frac{1}{n}$ is in $U$, so this implies that $y$ is not in $A$. Since $y$ was arbitrary, $x$ is an upper bound for $A$.$\square$

Let $x_1$ be an element of $(U-1)\setminus U$ (which is nonempty by Claim 1). Since $x_1$ is not in $U$, by Claim 2 there exists $n_1>1$ such that $x_1$ is not in $U-\frac{1}{n_1}$. Let $x_2$ be an element of $\left(U-\frac{1}{n_1}\right)\setminus U$. There exists $n_2>n_1$ such that $x_2$ is not in $U-\frac{1}{n_2}$, and we can take $x_3\in \left(U-\frac{1}{n_2}\right)\setminus U$; and so on. This yields an increasing sequence of positive integers $1=n_0<n_1<n_2<n_3<\cdots$ and an increasing sequence of real numbers $(x_k)$ such that $x_k$ is in $\left(U-\frac{1}{n_{k-1}}\right)\setminus U$. Then $(x_k)$ is bounded above by each element of $U$. Assuming MCT, $(x_k)$ converges to some real number $x$. Since $(x_k)$ is bounded above by each element of $U$ and $\lim x_k=x$, $x$ is less than or equal to each element of $U$. Since $x\geq x_k$ for all $k$, $x$ is in $\bigcap\limits_{j=1}^\infty\left(U-\frac{1}{n_j}\right)=\bigcap\limits_{n=1}^\infty\left(U-\frac{1}{n}\right)\subseteq U$. Therefore $x$ is the smallest element of $U$, which means $x=\sup A$.$\square$

Solution 2:

Contradiction is a very reasonable way to approach ($\Leftarrow$).

A hint for ($\Rightarrow$): Suppose that the sequence $\langle x_n:n\in\mathbb{N}\rangle$ is monotone increasing and bounded; then $\{x_n:n\in\mathbb{N}\}$ is bounded, so it has a least upper bound $u$. This $u$ looks like a very good candidate to be the limit of the sequence ... .