Weakly sequentially closed convex cone which is not weakly closed 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; cf. also this MO question 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.
 A: 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.
