Creating a nested sequence of compact subsets of an open set in $\mathbb{R}$

Question

Take $U \subset \mathbb{R}$ be an open set. I want to show that there exists nested compact sets $C_1 \subset C_2 \subset \dots$ such that $\bigcup_{j=1}^{\infty} C_j = U$. I have a feeling that I need to resort to the fact that $U$ can be formed by an infinite union of open intervals $I_j$, i.e. $U = \bigcup_{j=1}^{\infty} I_j$. I have a feeling for each $i \in \mathbb{N}$, I can form nested compact sets

$$C_{i1} \subset C_{i2} \subset \dots$$

such that $\bigcup_{j=1}^{\infty} C_{ij} = I_i$. Then I can set

$$C_i = \bigcup_{j=1}^{\infty} C_{ji}.$$

However, I am not sure if an infinite union of compact sets can potentially be compact. Hence, I could use some help on moving through this problem. Potentially I can do something involving the infinite intersection of compact sets?

Answer 1

What you're worried about can happen. Consider the open set $(0, \infty)\setminus\mathbb{N}$. It can be written as a the countable union of disjoint open intervals:

$$(0, \infty)\setminus\mathbb{N} = \bigcup_{n =1}^{\infty}(n-1, n).$$

Each open interval can be written as a union of nested compact sets:

$$(n - 1, n) = \bigcup_{k = 1}^{\infty}\left[n - 1 + \frac{1}{k + 1}, n - \frac{1}{k + 1}\right].$$

So, in your notation, we have $C_{nk} = \left[n - 1 + \dfrac{1}{k + 1}, n - \dfrac{1}{k + 1}\right]$. Then we have

$$C_k = \bigcup_{n = 1}^{\infty}C_{nk} = \left[\frac{1}{k}, 1 - \frac{1}{k}\right] \cup \left[1 + \frac{1}{k}, 2 - \frac{1}{k}\right] \cup \left[2 + \frac{1}{k}, 3 - \frac{1}{k}\right] \cup \dots$$

which is unbounded and hence not compact.

One way to rectify this issue, both for this example and in general, is to instead take $C_k = C_{1k}\cup\dots\cup C_{kk}$ which is compact as it is a finite union of compact sets. I'll leave it to you to check that it is nested and the union of the $C_k$ is the open set you desire.

Answer 2

You may use that $\mathbb R$ is $\sigma$-compact, that is, it is the union of countably many compact sets $K_n$ (e.g. $K_n=[-n,n]$). Also, every open set $U$ is $F_\sigma$, that is the union of countably many closed sets $F_i$. You may assume that $F_i$ is contained in $F_{i+1}$ (since you may replace $F_i$ with the union $F_1\cup F_2\cup\cdots\cup F_i$). Then $U$ is the increasing union of $K_n\cap F_n$.

One way to see that every open subset $U$ of the reals is $F_\sigma$, if $U=\mathbb R$ then it is the union of the $K_n=[-n,n]$. If $U\not=\mathbb R$, say $\emptyset\not=G=\mathbb R \setminus U$, then $G$ is closed and it is enough to show that $G$ is a $G_\delta$ set, that is, the intersection of countably many open sets. You could let $G_n$ to be the set of all points that are distance less than $\frac1n$ to $G$, then clearly $G_n$ is open and $G$ is the intersection of the $G_n$.

Answer 3

There are only countably many $C_{ij}$'s. Enumerate them $C^n$, $n\in\omega$. Now let $K_n=\bigcup _{k\leq n} C^k$ for $n\in\omega$. Then $K_n$ is compact and $U=\bigcup _{n\in \omega} K_n$.