Let $G$ be a compact group. When is the trivial subgroup, $1$, isolated in the set of all closed subgroups of $G$? I think that the isolated identity is called "No small subgroup". Either way, I haven't studied Lie theory. I am thinking about the question from a basic abstract algebra(my knowledge is Dummit and Foote) and Topology(Munkres) point of view.
I have checked that for both $S^1$ and $S^1 \times S^1$ the identity subgroup is not isolated. I have also checked that if I consider any cyclic subgroup $A$ in $G$ then $\bar A$ is also a subgroup. I have used the topological group theory along with the one below:
Let $G$ be a compact group. Write $1=1_G$ for the identity element of $G$. Let $S(G)=$all closed subgroups of $G$.
Topologize $S(G)$ so that a basic open neighborhood of $H$ in $S(G)$ has the form, for each open $U$ containing $1$: [B(H,U) = {K \in S(G) : K \subseteq H.U \text{ and } H \subseteq K.U}.]
So we can think of $U$ as being an $\epsilon$-ball around the identity of $G$. Then $H.U$ is the set of points that are within $\epsilon$ of $H$. Symmetrically, $K.U$ is the set of points that are within $\epsilon$ of $K$. Thus $K$ is in $B(H,U)$ iff $K$ is within $\epsilon$ of $H$ and $H$ is within $\epsilon$ of $K$. (Cf Hausdorff metric)
I don't want a detailed answer if you would like to write it. I appreciate that. If you give me some ideas and hints in steps. That will help me a lot.