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
\Bbb R
$\endgroup$