3

Theorem: Let $M$ be a Cauchy complete metric space. Suppose we have a sequence of non-empty closed sets $A_1 \supseteq A_2 \supseteq \cdots$ such that $\text{diam}(A_n) \rightarrow 0$. Then $\cap_n A_n$ is a singleton.

The proof is a straightforward application of the Axiom of Countable Choice. I'm interested in tracking how this invocation of Choice affects the rest of analysis. I have done some research of my own, but haven't been able to find much.

  1. What results in analysis are proven using this theorem?

  2. Can this theorem be proven using a weaker form of Choice? Or, if possible (but I highly doubt it), no Choice at all?

Hanul Jeon
  • 27,376
JMM
  • 1,096
  • For your point 1: I don't see how you could use the theorem to show that a given space is Cauchy complete. You can certainly use it to show that a given space is not Cauchy complete, although in the cases I can think of you can easily bypass the theorem by giving an explicit Cauchy sequence with no limit. For your point 2: you need someone who knows much more than me! – Rob Arthan Aug 15 '23 at 22:03
  • I reformulated part 1 just as you answered, realizing that it didn't make much sense. – JMM Aug 15 '23 at 22:04
  • 1
    General comment: without at least countable choice, much of analysis goes down the drain. Case in point: https://math.stackexchange.com/questions/126010/continuity-and-the-axiom-of-choice – PhoemueX Aug 16 '23 at 08:06

0 Answers0