Completeness of $\mathbb{R}$: equivalent definitions.

I'm looking for a reference for the equivalence between the least-upper-bound property and the Bolzano–Weierstrass theorem.

As a consequence of the least-upper-bound property, we can prove that every bounded monotone sequence of real numbers converges. Since every sequence of reals numbers has a monotone subsequence, we get the Bolzano–Weierstrass theorem.

Where can I find a proof for the converse?

Thanks.

1 Answer

We can try to prove it ourselves...

Suppose $S\subset \mathbb R$ has an upper bound. Pick an upper bound $a_1$ so that there is some $s\in S$ such that $|a_i-s| < 1$.

We know that we can do this because being unable to pick such an $a_1$ means that $\{x : |a-x|<1\}\cap S = \emptyset$ for all upper bounds $a$. But the archimedean property of $\mathbb{R}$ then forces $S=\emptyset$, since we could continue to replace $a$ with $a-\frac{1}{2}$.

EDIT: Note that this is not is not circular, since we are assuming Bolzano-Weierstrass, with which it is simple to prove the archimedean property. See the comments.

Since this is not a least upperbound, there is another upper bound $a_2 <a_1$ so that there is some $s \in S$ such that $|a_2-s|<\frac{1}{2}$.

Keep going to obtain a strictly decreasing sequence $a_1>a_2> a_3>\ldots$ of upper bounds of $S$ such that for each $n$ there is some $s\in S$ such that $|a_n-s|< \frac{1}{n}$. The same argument as above ensures that such a sequence exists.

Bolzano-Weierstrass tells us that this sequence has a convergent subsequence $\{b_n\}\subset \{a_n\}$.

Set $b = \lim_{n\to \infty} b_n$. The construction of our sequence makes it clear that for any $n>0$, there is some $s\in S$ such that $b < s +\frac{1}{n}$. Thus, $b$ is a least upper bound.

• However the archimedean property of $\Bbb R$ is proven (at least it was that way in the course I took) using LUB property. I imagine that archimedean property follows in a nice way from Bolzano-Weierstrass, but imo, this would need clarification. – Wojowu May 3 '15 at 13:58
• All I need is that $\mathbb N$ is unbounded in $\mathbb R$. This is easy to prove if we are assuming Bolzono-Weirstrass as $|n-m|\geq 1$, so there can't be a convergence subsequece of natural numbers. – WSL May 3 '15 at 14:14