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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?

Kiran
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    As it has 4 close votes, I'd like to add my unofficial vote against closing this question, as the OP has certainly provided the work on answering it: he did answer it! –  Jun 10 '14 at 05:33

2 Answers2

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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.

Kiran
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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.

  • I think the topological theorem you quoted still works if "$\mathcal A$ is simply ordered by inclusion" is weakened to "every finite subcollection of $\mathcal A$ is connected." Thus the OP's question reduces to whether, in a compact group, the intersection of two closed connected subgroups is connected. – bof Jun 07 '14 at 07:28
  • Right, as long as $\mathcal{A}'$ has the finite intersection property, the theorem goes through! It seems reasonable that if the answer is 'yes', this is a good place to use the group property. –  Jun 07 '14 at 08:21