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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 $\Rightarrow$.

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3 Answers 3

up vote 4 down vote accepted

"$\Leftarrow$": 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 $x=\sup A$.$\square$

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@mespebjidom: I don't know; I have no plans to do so, but I'll try to reply to specific questions about it if you have any. –  Jonas Meyer Sep 30 '13 at 2:09
I have a shorter proof. Please check it out and comment math.stackexchange.com/questions/509412/…. Thanks –  user87274 Sep 30 '13 at 3:55
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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 ... .

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Hey, can I have some help with using proof by contradiction in this link: math.stackexchange.com/questions/509412/…? –  user87274 Sep 30 '13 at 2:54
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Here's an attempt at a proof:

Proposition: Completeness Axiom $\iff$ Monotone Convergence Theorem



Assume for contradiction that Completeness Axiom $\not\Leftarrow$ MCT. That is, $x_n$ is monotone increasing and bounded sequence of real numbers with a finite limit. By our assumption, $\displaystyle\lim_{n \to \infty} x_n = u > \text{sup }x_n$. However, this is a direct contradiction of the Monotone Convergence Theorem since a monotone sequence converges to sup $a_k$ if $a_k$ is increasing. Therefore, the Completeness axiom $\Leftarrow$ Monotone Convergence Theorem


Suppose that the sequence $x_n$ is monotone increasing and bounded. Then $x_n$ is bounded, so it has a least upper bound $u \equiv \text{sup }x_n$ by the completeness axiom. Now for every $\epsilon > 0$, there exists an $N$ such that $N > u - \epsilon$, since otherwise $u- \epsilon$ is an upper bound of $x_n$, which contradicts to $u$ being the supremum $x_n$. Then, since $x_n$ is increasing, for all $n>N$ we have $| c - x_n | = c - x_n \leq c - x_N < \epsilon$. Hence by definition, the limit of $x_n = u$. Therefore, the Completeness axiom $\Rightarrow$ Monotone Convergence Theorem

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The proof of MCT$\Rightarrow$Completeness is incorrect. I do not understand the intended meaning, for example at "That is" and "By our assumption." What assumption? The negation of the completeness axiom, if that is what is intended, is the statement that there is a nonempty set that is bounded above but has no least upper bound. –  Jonas Meyer Dec 11 '11 at 5:07
In the proof of Completeness$\Rightarrow$MCT, you wrote $N>u-\epsilon$ where you mean $x_N>u-\epsilon$, and you wrote $c$ in some places where you meant $u$. –  Jonas Meyer Dec 11 '11 at 5:12
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