2
$\begingroup$

This question already has an answer here:

Cantor's Intersection Theorem states that "if $\{C_k\}$ is a sequence of non-empty, closed and bounded sets satisfying $C_1 \supset C_2 \supset C_3 \dots$, then $\bigcap_{n \ge 1} C_n$ is nonempty.

If the term "compact sets" is replaced by "closed sets", the statement is not true. It makes sense to me, but couldn't find such a counterexample for it.

$\endgroup$

marked as duplicate by user228113, Claude Leibovici, user91500, Paul Plummer, N. F. Taussig Jan 31 '16 at 9:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ If a counterexample exists, it'll have to make use of the difference between "closed" and "compact." This difference is boundedness. $\endgroup$ – user296602 Jan 31 '16 at 5:17
  • $\begingroup$ You can create an ellipsis in MathJax by using \dots. $\endgroup$ – Michael Albanese Jan 31 '16 at 5:18
  • $\begingroup$ yah I noticed that the integer set is closed but unbounded, but not sure it is a valid example..@T.Bongers $\endgroup$ – Maggie Mak Jan 31 '16 at 5:19
1
$\begingroup$

Consider the sequence $C_n = [n, \infty)$.

$\endgroup$
  • $\begingroup$ why the intersection of these sets is empty? Is it because we can't find such a number in every set? $\endgroup$ – Maggie Mak Jan 31 '16 at 5:25
  • 1
    $\begingroup$ There is no real number which belongs to $C_n$ for every $n$, so the intersection is empty. $\endgroup$ – Michael Albanese Jan 31 '16 at 5:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.