When is a sequentially closed cone, closed? The following question I also posed here, but still got no answer. Let $X$ be a locally convex, Hausdorff topological vector space and $C\subseteq X$ a convex cone, which is sequentially closed. What are criteria, that would imply that $C$ is closed (in the topology of X)?Are there also "testifyable" criteria?
 A: Since this is too long for a comment, I post it as an answer:
Sorry, but I do not agree with Peter Michor's answer. There are certainly better examples, but this the first I can remember: There are countable inductive limits $X=\lim\limits_{\to} X_n$ of Frechet spaces which are not Hausdorff (take a decreasing sequence of open connected sets $U_n$ in the complex plane with empty intersection and $X_n=H(U_n)$ the Frechet space of holomorphic functions on $U_n$ together with the injective restriction maps). $X$ is the a quotient of the direct sum $\bigoplus X_n$ which is certainly bornological and the kernel of the quotient map is sequentially closed
(because convergent sequences in the direct sum are located and convergent in some finite sum) but it is not closed because the quotient is not Hausdorff.
The situation is better for metrizable spaces (of course, this is trivial) as well as for so-called Silva spaces (also called LS or DFS-spaces, countable inductive limits of Banach spaces with compact inclusions): In these cases, sequentially closed subspaces are closed.

By 8.5.28 in the book of Bonet and Perez-Carreras, Barrelled Locally Convex Spaces,
even sequentially closed subsets of Silva spaces are closed.

Edit. A simpler example (but possibly less relevant for analytical applications) is the space $X=\mathbb R^I$ endowed with the product topology (point-wise convergence of functions $f:I \to \mathbb R$) if $I$ is uncountable and of moderate cardinality (e.g. $I=\mathbb R$). Then $X$ is bornological (due to the cardinality restriction) and
$L=\lbrace f\in X: \lbrace i\in I: f(i)\neq 0\rbrace \text{countable}\rbrace$ is sequentially closed and dense in $X$.
A: In ordered vector spaces  the question is restated as follows:
When does the Archimidean property imply that the positive cone of an ordered vector space is closed? So that should help in your search.
Here is a simple example of a  condition involving only the cone.
Suppose that $X$ is a topological vector space with convex cone (wedge) $X_+$ . Suppose that there is $e\in X_+$ that is an order unit. That is, for each $x\in X$ there is $\lambda >0$ satisfying 
$$
x\in \lambda e -X_+
$$
In this case $X_+$ is sequentially closed if and only if it is closed. This is because $e$ must be an internal point of $X_+$. Supposing that $X_+$ is sequentially closed; if $x\in \overline{X_+}$, then $\alpha e + (1-\alpha) x$
is an internal point of $X_+$ for all $\alpha\in (0,1)$ because $X_+$ is convex. In particular
$$
\frac{1}{n+1} e + (1- \frac{1}{n+1}) x 
$$
is an internal point of $X_+$. Thus, $x$ is in $X_+$. 
A: If $X$ is bornological (carries the finest locally convex topology compatible with the given family of bounded sets, or the the given dual space), then sequentially closed implies closed.
Edit: As Jochen pointed out, this is wrong. Sorry.
