How to prove that the both definition of completeness of $\mathbb{R}$ are equivalent?

Question

In the definition of completeness of a set, in particular $\mathbb{R}$, I have seen the following definitions:

1. Dedekind: Every non-empty bounded of subset has a least upper bound (with respect to the natural order).
2. Cauchy: Every Cauchy sequence converges.

However, how can one prove that both of these definitions are equivalent ?

Answer 1

Proof of $(2)\Rightarrow(1)$: Let a nonempty bounded $A\subset{\mathbb R}$ be given. Then there is an $a\in A$ and an $M\in{\mathbb Z}$ with $M\geq x$ for all $x\in A$. This $M$ is an upper bound of $A$, and no $s<a$ is an upper bound of $A$. Define a sequence $(z_n)_{n\geq0}$ as follows:

$$z_n:=\min\left\{k\cdot2^{-n}\>\biggm|\>k\in{\mathbb Z}, \ k\cdot2^{-n}{\rm\ is\ an\ upper\ bound\ of\ }A\right\}\qquad(n\geq0)\ .$$ If $z_n=k_n\cdot2^{-n}$ then $(k_n-1)\cdot2^{-n}$ is no longer an upper bound of $A$. It follows that either $z_{n+1}=z_n$ or $z_{n+1}=z_n-2^{-(n+1)}$, hence $|z_n-z_{n+1}|\leq 2^{-(n+1)}$. By the triangle inequality this implies $$|z_m-z_n|<2^{-n}\qquad(m\geq n\geq0)\ .$$ Therefore $(z_n)_{n\geq0}$ is a Cauchy sequence in ${\mathbb R}$, and by assumption converges to a point $\xi\in{\mathbb R}$.

I claim that $\sup A=\xi$.

Proof. If there is an $a\in A$ with $a>\xi$ then there is an $n$ with $\xi<z_n<a$, contrary to the definition of $z_n$. It follows that $\xi$ is an upper bound of $A$. If there is a smaller upper bound $\sigma$ of $A$ then there is a binary rational $b=k\cdot2^{-n}$ with $\sigma<b<\xi\leq z_n$, again contradicting the definition of $z_n$. It follows that $\xi$ is in fact the smallest upper bound of $A$.$\quad\square$

Proof of $(1)\Rightarrow(2)$: From $(1)$ it immediately follows that monotone bounded real sequences are convergent.

Let a Cauchy sequence $(x_k)_{k\geq0}$ in ${\mathbb R}$ be given. It is well known that such a sequence is bounded. Define the numbers $$a_n:=\inf_{k\geq n}x_k,\quad b_n:=\sup_{k\geq n}x_k\qquad(n\geq0)\ .$$ Then $a_n\leq b_n$ for all $n$, the $a_n$ form a bounded increasing sequence with $\lim_{n\to\infty}a_n=\alpha$, and the $b_n$ form a bounded decreasing sequence with $\lim_{n\to\infty}b_n=\beta\geq\alpha$.

Let an $\epsilon>0$ be given. Then there is an $n$ with $x_l-x_k\leq \epsilon$ for all $k$, $l\geq n$. It follows that $$\beta-\alpha\leq b_n-a_n=\sup_{l\geq n}x_l-\inf_{k\geq n} x_k=\sup_{k,\>l\geq n}(x_l-x_k)\leq\epsilon\ .$$Since $\epsilon>0$ was arbitrary this implies that in fact $\alpha=\beta$. From $$a_n\leq x_k\leq b_n\qquad(k\geq n)$$ it then easily follows that $\lim_{k\to\infty} x_k=\alpha$ as well.$\quad\square$

(Of course these things are proven in most textbooks on real analysis.)

Answer 2

You can try to add another step in between which is the famous

Bolzano-Weierstrass Theorem: Every infinite bounded set has an accumulation point.

And then prove them in cyclic fashion : Dedekind implies Bolzano implies Cauchy implies Dedekind. The first step can be completed by starting with an infinite bounded set $S$ and another set $$A=\{x\mid x\in\mathbb{R}, x\text{ exceeds only a finite number of members of }S \} $$ (note that $0$ is also "a finite number" so that $x\in A$ if $x$ does not exceed any member of $S$).

Then $A$ is bounded above (any upper bound of $S$ is also an upper bound of $A$) and $M=\sup A$ exists according to Dedekind's version of completeness. Consider the interval $(M-\epsilon, M+\epsilon) $ for any arbitrary $\epsilon>0$. There is a member $a\in A$ such that $a>M-\epsilon$ and hence only a finite number of members of $S$ are less than $a$. On the other hand since $M+\epsilon$ is not in $A$ so there are an infinite number of members of $S$ which are less than $M+\epsilon$ and out of these only a finite number are less than $a$. It follows that there are infinitely many members of $S$ in $(M-\epsilon, M+\epsilon)$. Thus $M$ is an accumulation point of $S$.

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Next we establish Bolzano-Weierstrass implies Cauchy. Let $\{x_{n}\} $ be a Cauchy sequence then it is bounded. And if the sequence takes only a finite number of values it is easy to see that it becomes constant after a certain point and converges to the same constant. So let the range of the sequence $\{x_{n} \}$ be infinite. Then by Bolzano-Weierstrass the sequence has an accumulation point $x$. We show that $x_{n} \to x$ as $n\to\infty$. Let $\epsilon>0$ be arbitrary. There is a positive integer $N$ such that $|x_{m}-x_{n} |<\epsilon /2$ whenever both $m, n$ exceed $N$. And since $x$ is an accumulation point of the sequence $\{x_{n} \} $ it follows that there is a value of $m>N$ such that $|x_{m} - x|<\epsilon/2$. Therefore we have $$|x_{n} - x|\leq|x_{n} - x_{m} |+|x_{m} - x|<\epsilon $$ for all $n>N$ and this means that $x_{n} \to x$ as $n\to\infty$.

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Finally we establish that Cauchy implies Dedekind. Let $S$ be a non-empty set bounded above and $K$ be an upper bound for $S$. Let $a\in S$. We construct sequences $x_{n}, y_{n} $ in the following manner. Let $x_{1}=a, y_{1}=K$. If $(x_{n}+y_{n})/2$ is an upper bound for $S$ then $$x_{n+1}=x_{n},y_{n+1}=\frac{x_{n}+y_{n}} {2} $$ otherwise $$x_{n+1}=\frac{x_{n}+y_{n}} {2} ,y_{n+1}=y_{n}$$ This way the sequences are obtained in a recursive manner. It should be clear from the above construction that for every $n$ there is some member $s\in S$ (depending on $n$) such that $x_{n} \leq s$ and $y_{n}$ is an upper bound for $S$ for all values of $n$. Further it is clear that $x_{n}$ is non-decreasing and $y_{n} $ is non-increasing and $$x_{n} \leq y_{n}, y_{n} - x_{n} =\frac{y_{1}-x_{1}} {2^{n-1}} $$ Clearly this also implies that $|x_{n+1}-x_{n}|\leq (y_{1}-x_{1})/2^{n-1}$. It is easily proved that $x_{n} $ is a Cauchy sequence and let's say it converges to $x$. By the given relations between $x_{n}, y_{n} $ it follows that $y_{n}$ also converges to $x$.

We show that $x=\sup S$. Since $y_{n} $ is an upper bound for $S$ it follows that $y_{n} \geq s$ for any $s\in S$. It follows by taking limits that $x\geq s$ and hence $x$ is also an upper bound for $S$. Since $x_{n} \to x$ it follows that for any $\epsilon>0$ there is some $x_{n} >x-\epsilon$. And by construction there is an element $s\in S$ such that $s\geq x_{n} $. Thus $s>x-\epsilon$ and therefore $x=\sup S$.

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You may also try to prove equivalence of these formulations of completeness to gain better understanding.

Answer 3

Indeed 1. implies 2., but the converse does not hold for ordered fields in general unless the Archimedean Property is added as an extra hypothesis. This is the assumption implicitly made in "proofs" of that implication for claiming that the sequence constructed is Cauchy: the Archimedean Property is required for moving from some $\varepsilon>0$ to a natural number $n$ such that $n>\dfrac{1}{\varepsilon}$.

The field of formal Laurent series is an example of an ordered field that is Cauchy-complete but not Dedekind-complete. See

James Propp, Real analysis in reverse, The American Mathematical Monthly 120:5 (2013), 392-408

for more details on this (and also other equivalences and non-equivalences of Dedekind-completeness).