0
$\begingroup$

In my textbook there is a theorem that states that the following is equivalent to a metric space, $M$, being complete:

For any descending sequence $A_1 \supset A_2 \supset A_3 \supset \cdots$ of nonempty closed sets with diameters approaching 0, the intersection $\cap A_i$ is nonempty.

Would this definition of completeness not imply that all metric spaces are complete? If we let $A_1$ be a closed subset of $M$. Then we let $A_2 = \{a\}\subset A_1$. Here we have a sequence $A_1 , A_2$ where the diameter approaches (even becomes) 0, and $A_1\cap A_2\ne \emptyset$. Furthermore $A_2$ is also a closed set (right?). What have I missed?

  • 1
    In this case, "any" is supposed to mean "every". ​ ​ – 2017-02-19
  • 0
    It's a similar mistake to saying "every space $X$ has a finite cover $\{X\}$ so every space is compact". One example shows nothing here. It's about *all* decreasing sequences and *all* open covers. – 2017-02-19

1 Answers 1

2

Absolutely not. Consider the sequence of closed sets in $\mathbb{Q}$ defined by:

$$A_n=\Bigg[\sqrt{2}-\frac{1}{n},\sqrt{2}+\frac{1}{n}\Bigg]\cap\mathbb{Q}$$

These sets are closed in $\mathbb{Q}$, with diameter going to $0$, but their intersection is the empty set.

In fact, I think you have misenterpreted the definition, A metric space $M$ is complete if and only if every sequence with such properties have a nonempty intersection. The example you just gave is a simple case.