Lebesgue covering theorem

Question

I am having trouble understanding Lebesgue covering theorem as stated in Mathematical Encyclopedia.

1. First of all I think I have confusion with the definition of "finite subsystem". Is it finite number of subsets(sets containing intervals) or finite number of elements(intervals themselves)? I have taken the second interpretation.

2. So there can't be an infinite number of open intervals covering a closed interval on R?

Answer 1

1. If your original system $G$ is a collection of sets, say $G = \{G_i\}_{i\in I}$ for some (possibly infinite) index set $I$, then a subsystem $G'$ is simply a subset of the sets in $G$, i.e. $G' = \{G_i\}_{i\in J}$, where $J\subseteq I$. If $J$ is finite, then $G'$ is a finite subsystem. For example, the collection of open intervals $\{(n-1/2, n+1)\}_{n\in\mathbb{Z}_{\geq 0}}$ covers the interval $I = \left[0,1\right]$, but we can also cover $I$ with just $\{(-1/2, 1), (1/2, 2)\}$. So you have chosen the correct interpretation.
2. There can be an infinite number of open intervals covering a closed interval, but if the closed interval in question is bounded, then any infinite cover can be reduced to a finite subcover: so we can throw out infinitely many of the sets in our cover and still cover the closed bounded interval, like in the example above for $\left[0,1\right]$.

Answer 2

As to (1), it's a finite number of sets.  

As to (2), no. It says that if you cover a closed and bounded set $S$ by an infinite number of open sets (let's say open intervals $I_i$, which is enough as every open set is a collection of such intervals (in the standard topology of $\mathbb{R}$ you are working with)),  then you can always choose only a finite number of these intervals that still cover $S$ (not every finite subcollection - the theorem only says there is at least one). And this must work no matter how you choose the $I_i$, as long as $S \subset \cup_{i =1}^{\infty} I_i$; if you find one collection of covering intervals for which no finite covering collection exists, S is either not closed or bounded. 

I'll try to give you an intuition for why this doesn't work when a set is either not closed or not bounded.  

(a) Not bounded: let $S := \mathbb{Z}$, the integers. It is clearly not bounded, and is trivially covered by $\{I_i \}$ if we let $I_i := (i - \frac{1}{2}, i + \frac{1}{2})$. But note that $I_i \cap I_j = \emptyset$ for $i \neq j$, and that each $i \in S$ lies in exactly one such interval (by construction, $i \in I_i$ for all $i$).  So no finite sub collection of $\{ I_i \}$ will cover $S$. Intuitively, "$S$ escapes to $\infty$".

(b) Not closed:  Let $S := (0, 1)$, which is not closed, and cover it by $ \{ I_i := ( \frac{1}{n}, 1) \} := \hat{C}$ for $n \in \mathbb{N}$.  As for $n \rightarrow \infty$ the left interval end goes to $0$, the collection covers $S$: $ S \subset \cup_{i=1}^{\infty} I_i$. Now any finite subcollection $C \subset \hat{C}$  has a largest $n$ such that $I_n \in C$, but $I_j \not\in C$ for any $j > n$ (if not, it is still an infinite subcollection). Hence, $\cup_{I_i \in C} I_i = I_n = (\frac{1}{n}, 1)$, $\frac{1}{n} \in (0,1)$, and $(0, \frac{1}{n} ] \subset (0,1)$ is not covered by any such $C$. Intuitively, at its boundaries, "an open set is too fine to be covered" by all such infinite coverings.  Inversely, if $S$ includes its boundaries ($[0,1]$, say), then there is one $I_k$ that covers the boundary. $\hat{C}$ is not an infinite cover of $[0,1]$, but $\bar{C} := \hat{C} \cup (- \epsilon, \epsilon) := I_0 \cup (1 - \epsilon, 1+ \epsilon) := I_{-1}$ is an infinite cover of $[0,1]$. If we choose $n$ such that $\frac{1}{n} < \epsilon$, then, quite obviously, $[0,1] \subset \cup_{i=1}^n I_i \cup I_0 \cup I_{-1}$, and we have found a finite subcollection.   

Maybe this gives you some intuition. You can also google "Heine-Borel Theorem" for probably more hits.