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I tried proving that

Every bounded increasing sequence converges in $\mathbb{R}$.

implies that $\mathbb{R}$ has the least upper bound. Here, $\mathbb{R}$ is taken as an ordered field which contains the natural numbers.

I came to a conceptual problem in a particular step of my atempt, which I don't know if it is justified. I'll highlight it below:


Proof: Since you can prove the Archimedean property directly from MCT, it also follows easily the density of rationals.

(IDEA: We take an increasing sequence. We would want it to converge to the $\sup$, but it may not necessarily converge (indeed, it most likely will not). We proceed with stubbornness.)

Let $A$ be bounded by above, non-empty. Assume that there is no least upper bound of $A$.

Consider $\Omega$ (the first uncountable ordinal).

Define a function $f: \Omega \rightarrow \mathbb{Q}$ by transfinite recursion:

Take $f(0)$ to be any rational smaller than an element of $A$.

Given $f(a)$, take $f(a+1)$ to be a rational greater than $f(a)$ and smaller than an element of $A$.

$\color{red}{\text{Given a limit ordinal $\gamma$, suppose we have defined an increasing function on the ordinals}}$ $\color{red}{\text{for all ordinals smaller than $\gamma$}}$. We have that there exists an increasing sequence of ordinals $\alpha_n$ smaller than $\gamma$ that converges to the limit ordinal (order topology). The associated $f(\alpha_n)$ is a bounded increasing sequence of real numbers, hence converge to a given real number $x$. Take a rational number which is greater than $x$, and smaller than an element of $A$ (this is guaranteed by the assumption that there is no least upper bound of $A$).

We have thus constructed an injection from $\Omega$ to $\mathbb{Q}$. Since $\mathbb{Q}$ is enumerable and $\Omega$ isn't, we have a contradiction.


My conceptual doubt comes from the red part... I am assuming not only that I have defined the function on the smaller ordinals, but also that the function satisfies something I want. Is this valid? Why?

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  • $\begingroup$ You only need $f$ to be increasing and below an element of $A$. $\endgroup$
    – 喻 良
    Oct 23 '19 at 4:52
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Pursuing your idea, but from above:

First, note that your hypothesis implies: $$\text{Every bounded decreasing sequence converges in $\Bbb R$,}\tag{BDSC}$$ because if $(a_n)$ is a bounded decreasing sequence in $\Bbb R$ with limit $L$, then $(-a_n)$ is a bounded increasing sequence with $(-a_n)\to -L$.

It's also worth stating that $$ \text{If $(s_n)$ is a bounded decreasing sequence with $\lim_n s_n = L$, then $L = \inf_n s_n$,}$$

i.e. $L$ is the greatest lower bound of the $s_n$.

Suppose $A\ne \emptyset$ has an upper bound but no least upper bound. For $x\in \Bbb R$, let $A<x$ mean that $x$ is an upper bound of $A$: $a< x$ for all $a\in A$. Then by hypothesis, $$\text{if $A<x$ then there is $y<x$ such that $A<y$.}\tag{Hyp}$$

Define $f\colon\omega_1\to \Bbb R$ as follows:

  1. $A$ is bounded above by some $M$, so let $f(0) = M;$
  2. $f(\alpha + 1) =$ some $y$ such that $A<y$ and $y<f(\alpha)$, assuming it exists; if there is no such $y$, let $f(\alpha + 1) =$ an element of $A$ — as $A$ has no greatest element, such a value will not be an upper bound of $A$;
  3. for limit $\lambda$, let $(\alpha_n)_{n<\omega}$ be some strictly increasing sequence of ordinals with $\sup_{n}\alpha_n = \lambda$, and define $f(\lambda) = \lim_n f(\alpha_n)$ (which will equal $\inf_n f(\alpha_n))$, provided $n\mapsto f(\alpha_n)$ is a bounded decreasing sequence; otherwise, let $f(\lambda)$ be some element of $A$.

By induction,

  • for all $\alpha<\omega_1$, $A < f(\alpha).$

    This uses (Hyp) at successor stages $\alpha+1$ and (BDSC) at limit $\alpha$.

  • For all for all $\alpha,\beta <\omega_1$, if $\alpha<\beta$ then $f(\beta) < f(\alpha).$

So $f$ is an injection $\omega_1\to\Bbb R$. But there is no such $f$, otherwise we could choose a rational in each interval $\left(f(\alpha+1), f(\alpha)\right)$.

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I also came up a similar question and found a proof. Then I googled this. Here is my proof without using Axiom of Choice.

A ZF and transfinitely-inductive proof that Monotone convergence theorem implies Largest-lower-bound property

Theorem [$\mathbf{ZF}$] Monotone convergence theorem implies Largest-lower-bound property for sets of reals.

Proof: Fix a bijection $f$ of $\omega$ to $\{a\}\cup \mathbb{Q}$.

Let $A$ be a nonempty subset of an interval $[a, b]$. We prove that $A$ has the largest lower bound by a transfinite induction method. Let $x_0=a$. So $x_0$ is a lower bound of $A$. If $x_0$ is the largest low bound of $A$, then we are done. Otherwise, let $x_1\in (x_0,b)$ be a rational lower bound.

At any countable stage $\alpha$.

Case(1). $\alpha=\beta+1$ for some $\beta$. If $x_{\beta}$ is the largest lower bound of $A$, then we are done. Otherwise, let $x_{\alpha}\in (x_{\beta},b)$ be a rational lower bound.

Case(2). Otherwise, $A_{\alpha}=\{n\mid \exists \beta<\alpha(f(n)=x_{\beta})\}$. Define $\{n_i\}_{i\in \omega}$ to be the sequence from $A_{\alpha}$ so that for any $i$ and $m<n_i$ in $A_{\alpha}$, $f(m)=x_{\beta_0}$ and $f(n_i)=x_{\beta_1}$ implies $\beta_0<\beta_1$. Then the sequence $\alpha_i$ so that $f(n_i)=x_{\alpha_i}$ is a cofinal sequence of $\alpha$. By Monotone convergence theorem, let $z=\lim_{i}x_{\alpha_i}$. Then $z$ is a lower bound of $A$. If $z$ is the largest lower bound, then we are done. Otherwise, let $x_{\alpha}>z$ be a rational lower bound of $A$ and continue the construction.

Note that $x_{\beta}<x_{\alpha}$ for any $\beta<\alpha$ so that $x_{\alpha}$ exists. And $x_{\beta}$ is rational except only $\beta=0$. By the countability of rationals, the construction stops at some countable ordinal $\alpha$. By the construction, the largest lower bound of $A$ exists.

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