# Two questions about increasing unions of compact subsets of a locally compact Hausdorff group.

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!

• I’m assuming that a Haar measure has been fixed on $G$. – Berrick Caleb Fillmore May 11 '15 at 4:38

Alright! I have managed to find answers to my questions.

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.

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.