Let $V$ be the real vector space of finitely supported functions $f: \Omega\to \mathbf{R}$ such that $\sum\_\omega f(\omega)=0$, where $\Omega$ is a given uncountable set.


Endow $V$ with the weak topology $\sigma:=\sigma(V,\mathbf{R}^\Omega)$, so that, more explicitly, a net $(f\_i)$ in $V$ is $\sigma$-convergent to $f \in V$ iff $\sum\_\omega f\_i(\omega)g(\omega)\to \sum\_\omega f(\omega)g(\omega)$ for all $g: \Omega\to \mathbf{R}$.



> 
> **Question 1.** Does there exist a $\sigma$-sequentially-closed convex cone in $V$ which is not $\sigma$-closed?
> 
> 
> 


Of course, any real vector space $E$ with uncountable dimension can be seen, up to isomorphism, as the vector space of finitely supported functions $f:\Omega\to \mathbf{R}$ with $\mathrm{dim}(E)=|\Omega|$, and $\mathbf{R}^\Omega$ can be seen as the algebraic dual $E^\star$.
Hence, ignoring the additional constraint $\sum\_\omega f(\omega)=0$, we may ask:



> 
> **Question 2.** Let $E$ be a real vector space with uncountable dimension. Does there exist a $\sigma(E,E^\star)$-sequentially-closed convex cone in $V$ which is not $\sigma(E,E^\star)$-closed?
> 
> 
> 


Ps. I know that the answer to Question 1 is negative if $\Omega$ is countable, but such examples may exist if $\Omega$ is uncountable. A related MSE question with a beautiful answer can be found [here](https://math.stackexchange.com/questions/2134762/is-there-any-example-of-a-sequentially-closed-convex-cone-which-is-not-closed); cf. also this [MO question](https://mathoverflow.net/questions/234537/every-convex-sequentially-closed-set-is-closed) for subsets which are not necessarily cones.


Ps2. A related result (Treves, "Topological Vector Spaces, Distributions and Kernels" (1967), p.201, Exercise 19.1): "Let $E$ be a locally convex Hausdorff tvs over $\mathbf{C}$. Then every linear subspace of $E$ is $\sigma(E,E^\star)$-closed." In our case, let $E\_0$ be a vector subspace of a real vector space $E$, let $\mathscr{B}\_0$ be a basis of $E\_0$ and extend it to a basis $\mathscr{B}$ of $E$. Since every linear functional is $\sigma(E,E^\prime)$-continuous, then $E\_0$ can be written as intersection of $\sigma(E,E^\prime)$-closed sets $\{x=\sum\_{b \in \mathscr{B}}\lambda\_{x,b}b \in E: \lambda\_{x,b\_0}=0\}$ for each $b\_0 \in \mathscr{B}\setminus \mathscr{B}\_0$. This proves that, if an example in Question 2 exists, then the $\sigma(E,E^\prime)$-sequentially closed convex cone is not a subspace.



Weakly sequentially closed convex cone which is not weakly closed
The answer is **negative**, for both questions.


Let $\{Z\_1,Z\_2\}$ be a partition of $\Omega$ such that $|Z\_1|=|Z\_2|$.
Let also $h: Z\_1\to Z\_2$ be a bijection and fix $a \in Z\_1$.
For each nonempty finite subset $B\subseteq Z\_1$, define
$$
\tilde{e}(B):=\frac{1}{|B|^2}\sum\_{b \in B}(e\_b-e\_{h(b)}),
$$
where $e\_z\in V$ stands for the function such that $e\_z(t)=1$ if $t=z$ and $e\_z(t)=0$ otherwise.
Note that $\tilde{e}(B)$ belongs to $$\Sigma:=\{f \in V: f\ge 0 \text{ and }\sum\nolimits\_\omega f(\omega)=1\}$$ for each finite nonempty subset of $Z\_1$. Now, let $\mathscr{B}$ be the family of nonempty finite subsets of $Z\_1\setminus \{a\}$ and $C$ be *the sequential weak-closure of the convex cone* $C\_0$, where
$$
C\_0:=\mathrm{co}\left(\mathrm{cone}\left(\left\{\tilde{e}(\{a\})+\tilde{e}(B): B \in \mathscr{B}\right\}\right)\right).
$$
(Here, $\mathrm{co}$ is the convex hull and $\mathrm{cone}(S):=\{\lambda x: \lambda>0, x \in S\}$.)
It is routine to check that $C$ is, indeed, a convex cone. To complete the proof, it will be enough to show that $\tilde{e}(\{a\}) \notin C$ and, on the other hand, $\tilde{e}(\{a\})$ belongs to the weak-closure of $C\_0$, so that $C$ is a sequentially weak-closed convex cone in $\Sigma$ which is not weak-closed.


$\text{ }$


$\bullet$ First, let us show that $\tilde{e}(\{a\}) \notin C$. For the sake of contradiction, suppose that, for each $n\ge 1$, there exist a positive integer $k\_n$, positive scalars $\lambda\_{n,1},\ldots,\lambda\_{n,k\_n} \in \mathbf{R}$, and
nonempty finite pairwise disjoint
sets $B\_{n,1},\ldots,B\_{n,k\_n}\in \mathscr{B}$ such that the sequence $(p\_n)\_{n\ge 1}$ is weak-convergent to $\tilde{e}(\{a\})$, where
$$
p\_n:=\sum\_{i=1}^{k\_n}\lambda\_{n,i} \left(\tilde{e}(\{a\})+\tilde{e}(B\_{n,i})\right).
$$
Set also $B\_0:=\bigcup\_{n\ge 1}\bigcup\_{i=1}^{k\_n}B\_{n,i}$.
Recalling that $\lim\_n \langle p\_n,u\rangle =\langle \tilde{e}(\{a\}),u\rangle$ for each $u \in V$ and choosing $u=e\_a$, we obtain $\lim\_n \sum\_{i=1}^{k\_n}\lambda\_{n,i}=1$; here, $\langle f,g\rangle:=\sum\nolimits\_\omega f(\omega)g(\omega)$ for each $f \in C$ and $g \in \mathbf{R}^\Omega$. First, suppose that $|B\_0|<\infty$ and write each $p\_n$ as $$\sum\_{\emptyset\neq B\subseteq B\_0}\lambda\_{n,B} \left(\tilde{e}(\{a\})+\tilde{e}(B)\right),$$ where each $\lambda\_{n,B}$ is possibly $0$ and at least one of them is nonzero. Letting $u$ be the characteristic function of $Z\_1$, we obtain
$$
1=\langle \tilde{e}(\{a\}), u\rangle
=\lim\_{n\to \infty}\langle \sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}\left(\tilde{e}(\{a\})+\tilde{e}(B)\right),u\rangle
=\lim\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}\left(1+\frac{1}{|B|}\right).
$$
Since $\lim\_n\sum\_{\emptyset \neq B\subseteq B\_0}\lambda\_{n,B}=1$, we obtain that
$$
\lim\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B|}=0. 
$$
However, this is impossible since
$$
\liminf\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B|} \ge 
\liminf\_{n\to \infty} \sum\_{\emptyset \neq B\subseteq B\_0}\frac{\lambda\_{n,B}}{|B\_0|}=\frac{1}{|B\_0|}>0.
$$
This contradiction proves that $B\_0$ is an infinite set. Hence, setting $z\_n:=\max \bigcup\_{i=1}^{k\_n}B\_{n,i}$ for each $n\ge 1$, we obtain that there exists a strictly increasing subsequence $(z\_{n\_m})\_{m\ge 1}$ with the property that $z\_{n\_m}\notin \bigcup\_{1\le t<n\_m}\bigcup\_{i=1}^{k\_t}B\_{t,i}$ for each $m\ge 1$. Finally, let $u\_0 \in \mathbf{R}^\Omega$ be the function supported on $\{z\_{n\_t}: t\ge 1\}$ defined by
$$
u\_0(z\_{n\_m}):=m \cdot \frac{\max\{|B\_{n\_m,i}|^2: i \in [1,k\_{n\_m}]\}}{\min\{\lambda\_{n\_m,i}: i \in [1,k\_{n\_m}]\}}
$$
for all $m\ge 1$. It follows by construction that
$$
\langle p\_{n\_m},u\_0\rangle\ge p\_{n\_m}(z\_{n\_m})u\_0(z\_{n\_m})
= u\_0(z\_{n\_m}) \sum\_{i=1}^{k\_{n\_m}}\lambda\_{n\_m,i} \left(\tilde{e}(\{a\})+\tilde{e}(B\_{n\_m,i})\right)(z\_{n\_m})
\ge u\_0(z\_{n\_m}) \sum\_{i=1}^{k\_{n\_m}}\lambda\_{n\_m,i} \tilde{e}(B\_{n\_m,i})(z\_{n\_m})
\ge u\_0(z\_{n\_m}) \cdot \frac{\min\{\lambda\_{n\_m,i}: i \in [1,k\_{n\_m}]\}}{\max\{|B\_{n\_m,i}|^2: i \in [1,k\_{n\_m}]\}}=m,
$$
for all $m\ge 1$. This shows that the subsequence $\left(\langle p\_{n\_m}, u\_0\rangle \right)\_{m\ge 1}$ cannot be convergent to $\langle \tilde{e}(\{a\}), u\_0\rangle$, contradicting the hypothesis that $(p\_n)\_{n\ge 1}$ is weak-convergent to $\tilde{e}(\{a\})$. Therefore $\tilde{e}(\{a\})\notin C$.


$\text{ }$


$\bullet$ Lastly, let us show that $\tilde{e}(\{a\})$ belongs to the weak-closure $\overline{C\_0}$ of $C\_0$. To this aim, suppose for the sake of contradiction that $\tilde{e}(\{a\})\notin \overline{C\_0}$. Thanks to the
Strong Separating Hyperplane Theorem, there exists a linear functional $f \in V$ such that
$$
f(\tilde{e}(\{a\}))=-1
\quad \text{ and }\quad 
f(p) \ge 0 \text{ for all }p \in \overline{C\_0}.
$$
Now, since $\Omega$ is uncountable, there exists a positive integer $k\_0$ and an uncountable subset $\tilde{Z}\_1\subseteq Z\_1\setminus \{a\}$ such that
$$
\forall b \in \tilde{Z}\_1, \quad f(\tilde{e}(\{a\})+\tilde{e}(\{b\})) \le k\_0,
$$
Note that, since $\tilde{e}(\{a\})+\tilde{e}(B)$ belongs to $C\_0$ for each $B \in \mathscr{B}$, then $f(\tilde{e}(B)) \ge 
1$.
To conclude, let $B$ be a finite subset of $\tilde{Z}\_1$ such that $|B|=k\_0+2$. It follows that
$$
0\le f(\tilde{e}(\{a\})+\tilde{e}(B))
$$
$$
=f\left(\frac{1}{|B|}\sum\_{b \in B}(\tilde{e}(\{a\})+\tilde{e}(\{b\}))+\tilde{e}(B)-\frac{1}{|B|}\sum\_{b \in B}\tilde{e}(\{b\})\right)
$$
$$
=\frac{1}{|B|}\sum\_{b \in B}f\left(\tilde{e}(\{a\})+\tilde{e}(\{b\})\right)+\left(\frac{1}{|B|^2}-\frac{1}{|B|}\right)\sum\_{b \in B}f\left(\tilde{e}(\{b\})\right)
$$
$$
\le \frac{k\_0}{|B|}+|B|\left(\frac{1}{|B|^2}-\frac{1}{|B|}\right)
<0.
$$
This contradiction proves that $\tilde{e}(\{a\})\in \overline{C\_0}$, completing the proof.


