Does weakly sequentially closed imply weakly closed? As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply  closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated. 
 A: The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of 
@mechanodroid Example. 
The converse is true in finite-dimensional Banach spaces or when the set considered is convex. 
A: It isn't true in general.
Let $H$ be an infinite-dimensional Hilbert space and let $(e_n)_n$ be an orthonormal sequence in $H$. Consider $S = \{\sqrt{n}e_n : n \in \mathbb{N}\} \subseteq H$.
$S$ is weakly sequentially closed because every weakly convergent sequence in $S$ has to be bounded, so the sequence has to have at most finitely many distinct terms, and hence it stabilizes to an element of $S$.
On the other hand, $0$ is in the weak closure of $S$. Indeed, consider a basic weakly open neighbourhood $U(0, a_1,\ldots, a_k, \varepsilon) = \{x \in H : |\langle x, a_i\rangle| < \varepsilon, 1 \le i \le k\}$ with $a_1, \ldots, a_k \in H$ and $\varepsilon > 0$.
Since $\sum_{n=1}^\infty |\langle a_i, e_n\rangle|^2 \le \|a_i\| < +\infty$, the series $\sum_{n=1}^\infty\sum_{i=1}^k |\langle a_i, e_n\rangle|^2$ also converges so there exists $n\in\mathbb{N}$ such that 
$$\sum_{i=1}^k |\langle a_i, e_n\rangle|^2 < \frac{\varepsilon^2}n$$
(otherwise we would have $\sum_{n=1}^\infty\sum_{i=1}^k |\langle a_i, e_n\rangle|^2 \ge \sum_{n=1}^\infty \frac{\varepsilon^2}n = +\infty$.)
For this $n$ we have $|\langle a_i, e_n\rangle| < \frac{\varepsilon}{\sqrt{n}}, \forall 1 \le i \le k$ so $\sqrt{n}e_n \in U(0, a_1,\ldots, a_k, \varepsilon)$.
Therefore every basic weakly open neighbourhood of $0$ intersects $S$ so $0 \in \overline{S}^w$. We conclude that $S$ is not weakly closed.
