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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

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  • $\begingroup$ You can use a \mathbb before R to get $\mathbb R $ $\endgroup$
    – gary
    Apr 9, 2015 at 4:52
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    $\begingroup$ @gary or \Bbb R $\endgroup$ Apr 9, 2015 at 5:20
  • $\begingroup$ @MarioCarneiro: Thanks, did not know that one. $\endgroup$
    – gary
    Apr 9, 2015 at 5:21

1 Answer 1

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

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  • $\begingroup$ No. So I can just let $x_n = n$ and say $x_n $ does not have a convergent subsequence? $\endgroup$
    – user848204
    Apr 9, 2015 at 4:40
  • $\begingroup$ Yes, that would work, since it is a sequence without a convergent subsequence. $\endgroup$
    – gary
    Apr 9, 2015 at 4:50

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