I am reading "Introduction to Set Theory and Topology" (in Japanese) by Kazuo Matsuzaka.
The definition of "sequentially compact" in this book is like the following:
Let $S$ be a metric space.
$S$ is said to be sequentially compact if every sequence of points of $S$ has a convergent subsequence.
The author proved the following theorem:
Theorem 12 on p.266:
Let $S$ be a metric space.
$S$ is compact if and only if $S$ is sequentially compact.
For example, let $S=(0,1)$.
Let $(a_n)$ be an arbitrary sequence in $S$.
Then, $(a_n)$ is a bounded sequence.
So, by Bolzano - Weierstrass Theorem in Calculus, $(a_n)$ has a convergent subsequence.
So, $(a_n)$ is sequentially compact by the author's definition.
So, by Theorem 12, $S=(0,1)$ is compact.
But this is not correct.
So, I think the following definition is correct:
Let $S$ be a metric space.
$S$ is said to be sequentially compact if every sequence of points of $S$ has a convergent subsequence which converges to a point in $S$.
Is the author's definition of "sequentially compact" ok?
Or do we need to modify it?