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The Setting

Let $K$ be an Archimedean field. TFAE:

  1. $K$ has the least upper bound property.
  2. Every Cauchy sequence in $K$'s additive group converges.

Now proving that 1 implies 2 is easy, but the other direction is slightly harder. Not that that's a problem. Rather the problem is that I can't see a route that doesn't invoke at least dependent choice at some point.

Strategy 1

Starting with a nonempty set $A$ that's bounded above, you could construct a monotonely non-decreasing Cauchy sequence of upper bounds that has the supremum as its limit. Here's a short sketch: Pick an upper bound $B_0$ of $A$. Pick an $a_0 \in A$. Recursively define $$ B_{i+1} = \begin{cases} \frac{B_i+a_i}{2}, & \text{ if that's an upper bound for } A \\ B_i, & \text{ otherwise} \end{cases} $$ and $$ a_{i+1} = \begin{cases} a_i, & \text{ if $\tfrac{a_i+B_i}{2}$ is an upper bound for $A$}\\ \text{choose any } a \in A \text{ s.t. } \frac{a_i+B_i}{2} < a, & \text{ otherwise.} \end{cases} $$ I can't see a way to get rid of the choice because you really want the $a_i$ to be in $A$ for the argument to go through.

Strategy 2

Okay, let's go the long way instead! First we show that $K$ complete implies $[a,b]$ compact. Then we show that that implies that closed and bounded subsets are compact ("Heine-Borel property"). And finally we show that not(Heine-Borel property) implies not(least upper bound property).

But I already get stumped on the first part. Clearly it's easy to show that $K$ complete implies $[a,b]$ is sequentially compact. And from here it'd be nice to use that $K$ is 2nd countable (the intervals with rational endpoints are a basis) to get that $[a,b]$ is in fact compact. So you start with an open covering $U_\alpha$ of $[a,b]$. 2nd countable spaces are Lindelöf... wait... let's make sure and prove that. Let $\{B_i\}$ be a countable basis. Then for each $B_i$ you choose a $U_\alpha$... oh. Choice crept up again.

So my question is this: Does this really require (an admittedly weak form of) choice? Or is there a way to do without?

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1 Answer 1

up vote 4 down vote accepted

It seems to me that you can modify strategy 1 as follows. Given a nonempty set $A$ that's bounded above, let $x_0$ be a non-upper-bound for $A$ and let $y_0$ be an upper bound for $A$. Recursively define $$ y_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 & \text{if this is an upper bound for $A$} \\ y_n & \text{otherwise.} \end{cases} $$ and $$ x_{n+1} \;=\; \begin{cases}(x_n+y_n)/2 & \text{if this is a non-upper-bound for $A$} \\ x_n & \text{otherwise.} \end{cases} $$ Note that $x_1 \leq x_2 \leq \cdots \leq y_2 \leq y_1$. Furthermore, each $y_n$ is an upper bound for $A$, and each $x_n$ is a non-upper-bound for $A$. It is easy to prove that both sequences are Cauchy sequences, and that they converge to the same limit $L$. We claim that $L$ is the least upper bound for $A$.

Both directions are fairly easy. If $a \in A$, then $a \leq y_n$ for all $n$, which proves that $a \leq L$. Thus $L$ is in fact an upper bound for $A$. Next, if $u$ is any upper bound for $A$, then $u \geq x_n$ for all $n$, and therefore $u \geq L$. Thus $L$ is the least upper bound for $A$.

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Yes, that seems correct. You use induction to show that $0 \leq y_i - x_i \leq 2^{-i}(y_0-x_0)$ and use that to get a bound on $|y_n - y_m|$. –  kahen Oct 15 '10 at 21:20

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