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

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?

• You can use a \mathbb before R to get $\mathbb R$ – gary Apr 9 '15 at 4:52
• @gary or \Bbb R – Mario Carneiro Apr 9 '15 at 5:20
Take the sequence {$1,2,3,...$} in $\mathbb R$ . Does it have any convergent subsequence?
• No. So I can just let $x_n = n$ and say $x_n$ does not have a convergent subsequence? – user848204 Apr 9 '15 at 4:40