Is an abelian group with a locally compact, 2-countable and Hausdorff topology such that the graph of addition and inverse are closed a Top group? Let $(G,+)$ be an abelian group endowed with a locally compact, second-countable and Hausdorff topology (but not necessarily a topological group). Suppose that, with respect to the product topology, the set $\{(g,h,g+h)\in G^3: g,h\in G\}$ is a closed subset of $G^3$ and the set $\{(g,-g)\in G^2:g\in G\}$ is a closed subset of $G^2$. Is it true that (with respect to that topology) $G$ is a topological group?
It is clear that assuming that $G$ is a topological group those sets are closed. But is the converse true?
I have found counterexamples in the case when we drop some of the hypothesis on the topology on $G$. For example, the Sorgenfrey line https://topospaces.subwiki.org/wiki/Sorgenfrey_line equipped with addition as operation. But I did not find any proof or counterexample if we restrict to the aforementioned setting.
 A: Let $G=\mathbb{Z}$ and give it the finest topology such that the sequence $(n^2)$ converges to $0$.  (Explicitly, every point except $0$ is isolated, and a neighborhood of $0$ must contain $n^2$ for all sufficiently large $n$; this space is homeomorphic to a disjoint union of a countably infinite discrete space and the one-point compactification of a countably infinite discrete space.)  I claim that addition and negation have closed graphs with respect for this topology.
For addition, suppose we have sequences $(x_n),(y_n)$ and $x,y,z\in\mathbb{Z}$ such that $x_n\to x$, $y_n\to y$, and $x_n+y_n\to z$; we must show $z=x+y$.  If $(x_n)$ and $(y_n)$ are both eventually constant, this is trivial, so we may assume $(x_n)$ is not eventually constant; passing to a subsequence, this means we may assume $x_n=k_n^2$ for some strictly increasing sequence $(k_n)$ and $x=0$.  If $(y_n)$ is not eventually constant then again we can pass to a subsequence to assume it is an increasing sequence of squares and $y=0$.  Then $(x_n+y_n)$ is increasing and so it can only converge to $0$ so $z=0$, as desired.  Thus we may assume $(y_n)$ is eventually constant, so $x_n+y_n=k_n^2+y$ for $n$ sufficiently large.  Again, the only way this converge is if these values are eventually all squares, but this is impossible unless $y=0$ since the differences between consecutive squares are eventually greater than $|y|$.  So $y=0$ and again we get $z=0$ as desired.
For negation, the argument is similar but easier.
