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!? :(

• The definition of a weakly closed set cannot easily be given in terms of weakly convergent sequences. The set $\{ \sqrt{n} e_n \mid n \in \mathbb{N}\} \subset \ell^2$ is weakly sequentially closed, but $0$ is in its weak closure. Jun 20, 2013 at 0:41
• @Martin Good point. The weak closure of a set $A\subseteq X$ may contain elements which are not w-limits of sequences of $A$. I just think you need to consider the set $\{\sqrt{n+1}e_n\}_n$ otherwise $0$ is in the set. Jul 7, 2016 at 17:46

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 $$A$$.

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.

• @Amarus I am surprised at the number of up votes for the answer by Brian M. Scott because the question has nothing to do with weak topology. Read the second answer ingnore the first one. Jun 29, 2018 at 6:11

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.

• But $x_n \to x$ weakly does not imply $x_n \to x$ strongly, so you cannot use the strong sequential closedness of $A$. Aug 15, 2017 at 5:46