Does a compact Lie group have finitely many conjugacy classes of maximal Abelian Lie subgroups?

Question

Let $G$ be a compact Lie group. An Abelian Lie subgroup $A \leq G$ is a maximal Abelian Lie subgroup if, for any Abelian Lie subgroup $A'$ such that $A \leq A' \leq G$, then $A' = A$.

Of course any maximal torus of $G$ (there is only one, up to conjugacy classes) is a maximal Abelian Lie subgroup, but there are other ones too, for example the Klein 4-group in $\mathrm{SO}(3)$.

What I'm wondering is if the number of conjugacy classes of maximal Abelian Lie subgroups of any compact Lie group $G$ is always finite?

Answer 1

Yes it's true. It essentially follows from a lemma quoted in the linked answer by user Qayum Khan, which I quote verbatim:

Neighboring subgroups theorem [1942] Any compact subgroup $H$ of an arbitrary Lie group $G$ admits a neighborhood $O$ in $G$ such that any subgroup contained in $O$ is $G$-conjugate to a subgroup of $H$.

Now let by contradiction $(A_n)$ be a sequence of pairwise non-conjugate closed maximal abelian subgroups in a compact Lie group. Let $A$ be a limit point in the Hausdorff topology; this is a compact abelian subgroup. By the above result, for $n$ large enough $A_n$ is conjugate to a subgroup of $A$. By maximality, this means that for $n$ large enough, $A_n$ is conjugate to $A$. This contradicts the non-conjugation.