Given an ordered field $F$, the following two statements are equivalent:

  1. $F$ has the Least-Upper-Bound property.

  2. $F$ is Archimedean and $F$ is "sequentially complete"/"Cauchy complete" (all Cauchy sequences in $F$ converge).

For some reason I have been unable to find a proof of this result on the web.

  • $\begingroup$ from a Cauchy sequence $x_n$, construct two monotone adjacent sequences, one of them can be empty or finite, but then the limit of $x_n$ is the least upper bound / largest lower bound of the other $\endgroup$
    – reuns
    Oct 5, 2016 at 2:33

1 Answer 1


First we will establish some intermediate results:

Theorem 1: Every Cauchy sequence is bounded.

Proof: Take $\epsilon=1$, then there is $N\in\mathbb{N}$ such that

$$m, n \geq N\quad\implies |x_m-x_n|<1$$

So that $$|x_m|<|x_N|+1\quad\quad\forall\;m\geq N$$ therefore $$|x_n|\leq\max\lbrace |x_1|,\ldots,|x_N|,|x_N|+1\rbrace\quad\quad\forall\;n\in \mathbb{N}$$

Theorem 2: Every sequence has a monotone subsequence.

Proof: Lemma in Bolzano-Weierstrass

Theorem 3: Let $F$ be an ordered field with the LUB property, then every bounded and monotone sequence in $F$ is convergent.

Proof: WLOG lets assume the sequence is increasing. As it is bounded it has a supremum: $s=\sup_{n\in\mathbb{N}}\lbrace x_n\rbrace$ We claim that $s$ is the limit of the sequence.

Let $\epsilon>0$ then there is $N\in\mathbb{N}$ such that

$$s-\epsilon<x_N\leq x_{N+1}\leq \cdots \leq s$$

so that for any $n\geq N$ we have $$|x_n-s|<\epsilon$$

Which proves that $s$ is the limit.

To simplify discussion, we assume that the order field is $\mathbb{R}$.

For the first implication $(1)\implies (2)$:

To establish the Archimedean Property first note that the set of integers is not bounded above, if it were it would have a supremum. Let's call it $z$; then there would be an integer $n$ such that $z-1<n$ but then $z<n+1\in \mathbb{Z}$ This contradicts the fact that $z$ was the supremum.

Archimedean Property: For every real $x$ there is an integer $n$ such that $n>x$

Proof: If this is not the case, integers would be bounded by some $x$.

Theorems $(1)$, $(2)$ and $(3)$ above prove Cauchy completeness.

For the reverse implication $(2)\implies (1)$:

Theorem 4: Suppose $F$ has the Archimedean property, then every monotone bounded sequence is Cauchy.

Proof: WLOG take the sequence to be increasing. If $\lbrace x_n\rbrace_{n}$ is not Cauchy then there exists $\epsilon>0$ so that for every $N\in\mathbb{N}$

$$n>m\geq N\quad\implies x_n-x_m\geq \epsilon$$

We are going to extract a subsequence such that it is not bounded.

For $N=1$ choose ${n_1},\ {n_2}$ such that $x_{n_2}-x_{n_1}>\epsilon$ Now take $N^{\prime}>n_2$ and choose ${n_3},\ {n_4}$ such that $x_{n_4}-x_{n_3}>\epsilon$. Continue in this way to construct a subsequence. Note that there are infintely many differences greater than $\epsilon$, so by the Archimedean Principle the subsequence diverges. This contradicts the fact that the sequence was bounded.

Observe that if $F$ is also Cauchy complete then every monotone bounded sequence is convergent.

We have established:

Cauchy completeness+Archimedean Property $\implies$ Convergence of every monotone and bounded sequence

The answer in this post proves:

Convergence of every monotone and bounded sequence $\implies$ LUB Property


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