Is intersection of connected subgroups connected?

Question

Let $G$ be a compact group. If $A_{\alpha}$, $\alpha \in I$ is a family of closed connected subgroups in $G$, then is it true that $\bigcap_{\alpha \in I}A_{\alpha}$ is connected?

Answer 1

This is a partial answer to your question, which doesn't use the fact that $G$ or $A_\alpha$ are groups. Maybe this argument could be patched up to give your result for topological (or Lie, I'm not sure) groups, or maybe this partial result is all you need for your application.

Problem 26.11 in Munkres, Topology 2nd edition, reads:

Let $X$ be a compact Hausdorff space. Let $\mathcal{A}$ be a collection of closed, connected subsets of $X$ that is simply ordered by proper inclusion. Then $Y=\bigcap_{A\in\mathcal{A}} A$ is connected.

The only difference is that Munkres assumes the collection $\mathcal{A}=\{A_\alpha: \alpha\in I\}$ is simply ordered; not sure if that's a deal-breaker or not for this problem.

By the way, the solution to the restricted case from Munkres' book goes like this (as per the hint in his book, which is to take a separation $C\cup D=Y$ of $Y$, and choose disjoint open sets $U, V\subset X$ which contain $X$ and $Y$ respectively.

Now look at the new collection of sets, $\mathcal{A}'=\{A-(U\cup V): A\in\mathcal{A}\}$. On the one hand, $\bigcap_{A'\in\mathcal{A'}} A'=Y-(U\cup V)=\emptyset$. However, $A-(U\cup V)$ is a closed subset of $A$ minus an open one, thus a closed subset of $A$ and thus compact itself. So by the Cantor intersection theorem, $\bigcap_{A'\in\mathcal{A'}} A'$ is nonempty. Contradiction, thus $Y$ is connected.

Answer 2

I found that this need not be true. For example, if $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2/ \mathbb{Z}^2$ is the canonical projection from $\mathbb{R}^2$ on to the torus $\mathbb{T}^2=\mathbb{R}^2/ \mathbb{Z}^2$, then the images of the lines $y=2x$ and $y=4x$ under $\phi$ are closed connected subgroups of $\mathbb{T}^2$ but their intersection is not connected.