Weakly closed implies sequentially closed Another problem involving the weak topology:
Let $X$ be a normed space and $A \subset X$ weakly closed. Then $A$ is sequentially closed, that is: If $(x_n) \subset A$ and $x_n \xrightarrow{w}x$, then $x \in A$.
I know this characterisation is also used as definition of weakly closed. So I guess it should be easy to prove. Yet I have trouble doing so. ;(
I tried proving it directly and towards a contradiction without success. I'm afraid my problem is a lack of understanding of weak closedness. I know how weakly open sets are generated, but this doesn't give me a concrete representation them, or of a weakly closed subset.
However, I know that $A$ is also closed with respect to the norm of $X$. Weak convergence of $(x_n)$ to $x$ means $f(x_n) \rightarrow f(x)$ for every $f \in X^*$. But it doesn't give me any statement related to the norm convergence of $(x_n)$, at least not that I know of. So the norm-closedness of $A$ doesn't really help. I could also find out that norm closedness is not sufficient for closedness with respect to weak convergence - I think convexivity has to be added to make the implication valid, is that correct?
So of course I have to fail if I weak convergence with norm convergence alone.
I've tried to work with balls around $f(x)$, too (for a contradiction). But again, the continuousness of $f$ only gives me control over $|f(x_n)-f(x)|$ if I have some bound for $||x_n-x||$. And I want it the other way around. I have a feeling that this way is wrong because it would need some implication between weak and norm convergence that I know isn't there...
It makes me mad that the proof should be rather simple, yet I'm not able to do it. Some hint please!? :(
 A: HINT: If you understand the product topology on an arbitrary product of topological spaces, you can use that to get a better handle on the weak topology on $X$: like the weak topology in $X$, the product topology is an example of an initial topology.
The weak topology on $X$ is the coarsest topology on $X$ that makes all $f\in X'$ continuous. For each $f\in X'$ let $R_f$ be a copy of $\Bbb R$ with the usual topology. Then the map
$$\varphi:X\to\prod_{f\in X'}R_f:x\mapsto\langle f(x):f\in X'\rangle$$
is an embedding. Thus, we can use the usual base for the product topology to define a base for the weak topology on $X$:

Let $\tau$ be the topology on $\Bbb R$. For each finite $F\subseteq X'$ and $U:F\to\tau$ let $$B(F,U)=\{x\in X:f(x)\in U(f)\text{ for each }f\in F\}\;.$$ The family of all such sets $B(F,U)$ is a base for the weak topology on $X$.

Now fix $x\in X$. For each finite $F\subseteq X'$ and $\epsilon>0$ let $$B(F,\epsilon)=\{y\in X:|f(x)-f(y)|<\epsilon\text{ for each }f\in F\}\;;$$ the family of all such sets is a local nbhd base at $x$ in the weak topology. In particular, if $A\subseteq X$ is weakly closed, and $x\notin A$, then there are a finite $F\subseteq X'$ and an $\epsilon>0$ such that $B(F,\epsilon)\cap A=\varnothing$, i.e., such that 
$$\text{for each }y\in A\text{ there is an }f\in F\text{ such that }|f(x)-f(y)|\ge\epsilon\;.\tag{1}$$
Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $A$ that converges weakly to $x$. Use $(1)$ and the finiteness of $F$ to $x$ to get a contradiction.
A: Instead of saying "$A$ is sequentially closed", the conclusion of the problem really should say that $A$ is "weakly sequentially closed" or "sequentially weakly closed." 
The problem is asking to show that closed in the weak topology implies sequentially closed in the weak topology.  
A solution can be given that does not use the specifics of the weak topology at all. For any topological space, closed implies sequentially closed.  
Let $X$ be any topological space and let be $A$ any closed subset of $X$. We need to show that $A$ is sequentially closed. Let $x \in X$ and $(x_n) \subseteq A$ be such that $x_n \to x$. We need to show $x \in A$. Let $U$ be an arbitrary  open neighbourhood of $x$. Since $x_n \to x$, we have $x_n \in U$ for all but finitely many $n$, and so $U$ contains at least one point of $A$. Since $U$ was an arbitrary open neighbourhood of $x$, this proves $x$ is in the closure of $A$. Since $A$ is closed, $x$ is in $A$.
A: Can't you just say that since $A$ is weakly closed then $A$ is strongly closed, then $A$ is sequentially closed? Cause I think every weakly closed set is strongly closed and in metric spaces (we are in a normed space which means in a metric space) (strongly) being closed set is equivalent to being sequentially closed.
