Definition for "relatively sequentially compact" A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.
Let $X$ be a topological space and $A \subseteq X$.
I've seen two definitions for $A$ to be relatively sequentially compact in $X$:


*

*the closure $\overline{A}$ of $A$ in $X$ is sequentially compact, which means that every sequence in $\overline{A}$ has a convergent subsequence (with limit in $\overline{A}$).

*every sequence in $A$ has a convergent subsequence with limit in $\overline{A}$.


Clearly, 1 => 2.
Are these definitions equivalent? (I don't see how to reduce the sequence in $\overline{A}$ to a sequence in $A$, so it seems that they are not equivalent unless $X$ is something like a Fréchet-Urysohn in which points in the closure $\overline{A}$ can be approximated by a sequence in $A$. Then we could try to perform a diagonal argument.) If they are not equivalent, what is the right definition for abstract topological spaces $X$?

There may be contexts where the first definition is appropriate, but it does
seem somewhat pathological in that a sequentially compact subspace may not be
relatively sequentially compact. In this respect it  differs
from the second definition.
For example, take the Tychonoff plank 
$X = ([0, \omega_1] \times [0, \omega]) \setminus (\omega_1, \omega)$
and the subspace $A = [0, \omega_1) \times [0, \omega]$.
Then $A$ is sequentially compact, since any sequence is confined to a
subspace $[0, \alpha] \times [0, \omega]$ with $\alpha < \omega_1$, which
is compact and first countable. On the other hand $\overline{A} = X$,
which is not sequentially compact since $\{ (\omega_1, n) \}_{n=0}^\infty$
has no cluster point.
To decide which is the most useful definition would probably involve looking
at a large number of applications, which I don't have available. I could 
even imagine some use for a definition $1\frac12$: a subspace $A$ of a
topological space $X$ is relatively sequentially compact if there is a 
sequentially compact $B \subset X$ such that $A \subset B$. This is strictly
weaker than definition 1 and stronger than definition 2, although I don't
know if it is strictly stronger than definition 2.
