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I have two questions to ask related to my research.

Question 1. Let $ G $ be a locally compact Hausdorff group. Is it possible that $ G $ is the union of a chain of compact subsets (ordered by inclusion) of $ G $?

As is well-known, ‘chain’ cannot be replaced by ‘countable chain’ as there are examples of locally compact Hausdorff groups that are not $ \sigma $-compact.

Question 2. Let $ G $ be as before, and suppose that $ X \subseteq G $ is an open $ \sigma $-finite subset. Then is there an increasing sequence of compact subsets of $ G $ whose union contains $ X $?

Thanks!

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  • $\begingroup$ I’m assuming that a Haar measure has been fixed on $ G $. $\endgroup$ – Berrick Caleb Fillmore May 11 '15 at 4:38
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Alright! I have managed to find answers to my questions.


Answer to Question 1: No.

Equip $ (\Bbb{R},+) $ with the discrete topology $ \tau_{d} $. It is clearly a locally compact Hausdorff group, and a subset of $ \Bbb{R} $ is $ \tau_{d} $-compact if and only if it is finite. Suppose that $ \mathcal{C} $ is a chain of finite subsets of $ \Bbb{R} $ ordered by inclusion, and also that $ \bigcup \mathcal{C} = \Bbb{R} $. Then inductively construct an increasing sequence $ (F_{n})_{n \in \Bbb{N}} $ in $ \mathcal{C} $ as follows:

  • Let $ F_{1} $ be any element of $ \mathcal{C} $.
  • Suppose that $ F_{n} $ has been chosen. If each element of $ \mathcal{C} $ were a subset of $ F_{n} $, we would never get $ \bigcup \mathcal{C} = \Bbb{R} $ because $ F_{n} $ is finite. Hence, there exists an element of $ \mathcal{C} $ that properly contains $ F_{n} $, which we fix to be $ F_{n + 1} $.

Observe that although $ \displaystyle \bigcup_{n = 1}^{\infty} F_{n} $ is an infinite subset of $ \Bbb{R} $, it is only countably infinite. This implies that there is an element $ F $ of $ \mathcal{C} $ that contains $ \displaystyle \bigcup_{n = 1}^{\infty} F_{n} $, which is clearly a contradiction, for $ F $ must be finite.


Answer to Question 2: Yes.

By the answer in this post, there exists a $ \sigma $-compact open subgroup $ H $ of $ G $. Our claim is that $ X $ intersects at most countably many cosets of $ H $. Assume the contrary. Write $ X = \bigcup_{k = 1}^{\infty} X_{k} $, each $ X_{k} $ having finite measure. Then by the Infinite Pigeonhole Principle, $ X_{k} $ intersects uncountably many cosets of $ H $ for some $ k \in \Bbb{N} $. As the Haar measure $ \mu $ on $ G $ is regular, we can find an open subset $ U $ of $ G $ containing $ X $ such that $ \mu(U) < \infty $. However, we have the following:

  • $ U $ intersects uncountably many cosets of $ H $.
  • Each coset of $ H $ is an open subset of $ G $.
  • Every non-empty open subset of $ G $ has positive (possibly infinite) measure.

It follows immediately that $ \mu(U) = \infty $, which is a contradiction.

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