1
$\begingroup$

Show that $R$ is closed but not sequentially compact.

Attempt: A subset E of a metric space X is said to be sequentially compact if and only if every sequence $x_n \in E$ has a convergent subsequence whose limit belongs to $E$. And every sequentially set is closed and bounded.

Suppose $x_n$ is a sequence then $R$ is closed since every convergent sequence $x_k \in R$ satisfies $\lim_{k → \infty} x_n \in E$. But (0, 7) is bounded and closed but not sequentially compact?

Can someone please help? Any feedback will help. Thank you

Improve this question
$\endgroup$
  • $\begingroup$ You can use a \mathbb before R to get $\mathbb R $ $\endgroup$ – gary Apr 9 '15 at 4:52
  • 1
    $\begingroup$ @gary or \Bbb R $\endgroup$ – Mario Carneiro Apr 9 '15 at 5:20
  • $\begingroup$ @MarioCarneiro: Thanks, did not know that one. $\endgroup$ – gary Apr 9 '15 at 5:21
4
$\begingroup$

Take the sequence {$ 1,2,3,...$} in $\mathbb R$ . Does it have any convergent subsequence?

Improve this answer
$\endgroup$
  • $\begingroup$ No. So I can just let $x_n = n$ and say $x_n $ does not have a convergent subsequence? $\endgroup$ – user848204 Apr 9 '15 at 4:40
  • $\begingroup$ Yes, that would work, since it is a sequence without a convergent subsequence. $\endgroup$ – gary Apr 9 '15 at 4:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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