TITLE: What does "weakly compact" mean when applied to subsets $X \subset Y$?
QUESTION [6 upvotes]: Let $X$ be a subset of a Banach space $Y$. Please can you give me a definition of what "$X$ is weakly compact" means? I want one which is in terms of sequences and boundedness, as opposed to one with topology and stuff like that. Thank you. 
I have searched the internet for days to avail for such a nice definition.
I did receive an answer in this thread, however the answer makes no reference to the set $Y$. Also a citation to a text would be nice.

REPLY [8 votes]: As said in comments,  "$X$ is weakly compact" means "If we endow $Y$ with the weak topology, and consider its restriction to $X$, then $X$ is compact."  
Since the weak topology is not metrizable, we cannot completely describe it using sequences; e.g., we don't get a sequential characterization of the closure of a set. However, we can describe compactness in this topology  using sequences. This is what Eberlein–Šmulian theorem theorem says: a set $X\subset Y$ is weakly compact if and only if every sequence of elements of $X$ has a weakly convergent subsequence. This applies to any Banach space $Y$, reflexive or not. 
For the weak* topology we do not have an analog of Eberlein–Šmulian theorem: for example, the unit ball of $\ell_\infty = \ell_1^*$ is compact in the weak* topology, but the sequence of "standard basis" vectors $e_n = (0,0,\dots, 0,1,0,\dots)$ in $\ell_\infty$ has no weak* convergent subsequence. 
If $Y$ is reflexive, then the weak topology on $Y$ is also the weak* topology, considering $Y= (Y^*)^*$. So you get the best of both worlds: Banach-Alaoglu theorem provides compactness, while  Eberlein–Šmulian theorem provides a  description of that compactness in terms of sequences.
Recommended text: A Course in Functional Analysis by John B. Conway, Chapter 5.