What does the Heine-Borel Theorem mean?

Question

Question: What is the Heine-Borel Theorem saying?

I am working through Hardy's Course of Pure Mathematics (ed.3) and am attempting to understand what the Heine-Borel means (s. V, pp. 186).

Theorem: Suppose that we are given an interval $(a,b)$ and a set of intervals, $I$ each of whose members is included in $(a,b)$. Suppose further that $I$ posses the following properties:

(1) every point of $(a,b)$, other than $a$ and $b$, lies inside at least one interval of $I$.

(2) $a$ is the left-hand end point, and $b$ the right-hand end point, of at least one interval of $I$.

Then it is possible to choose a finite number of intervals from the set $I$ which form a set of intervals possessing the properties (1) and (2).

I am confused as to what this theorem is actually saying.

My interpretation is that there is some arbitrary interval consisting of real numbers, $(a,b)$ and a set of intervals, $I$, and if you were to take all the members of each set in $I$ out of it, and make a set of those, that set would include the exact points of $(a,b)$. Then satisfying those properties, $I$ can be reduced to some intervals which have the same characteristic I just described of it.

But based on the properties, couldn't I just take the interval $(a,b)$ in $I$ and then I got my finite number of intervals, $1$. My instincts tells me no, but I am unsure of why.

Answer 1

I was trying to learn analysis from Hardy a while ago, and I found it quite difficult due to his old fashioned way of speaking. In modern terms, he is saying that for intervals on the real line of type $[a,b]$, every open cover has a finite subcover. This is by no means a modern statment of the Heine-Borel Theorem. I'll disect what Hardy discusses below.

Definition. Let $A \subset \mathbf{R}$. An open cover of $A$ is a collection of sets $\{C_i\}_{i \in I}$ such that $A \subset \bigcup_{i \in I} C_i$. A subcover of $A$ is a subset of our collection of $C_i$s, i.e. a collection $\{C_j\}_{i \in J}$ where $J \subset I$ and $A \subset \bigcup_{j \in J} C_j$ .

Definition. Let $A \subset \mathbf{R}$. $A$ is compact if every sequence in $A$ has a convergent subsequence.

Property 1 (Definition). Let $A \subset \mathbf{R}$. $A$ satisfies property 1 if every open cover of $A$ has a finite subcover.

Property 2 (Definition). Let $A \subset \mathbf{R}$. $A$ satisfies property 2 if $A$ is closed and bounded. (Note that we say a set of real numbers is closed if every convergent sequence in that set has its limit in that set).

I have seen different statements of the Heine-Borel theorem, but here is one that encapsulates all of what it could possibly mean.

Theorem (Heine-Borel). $A \subset \mathbf{R}$ is compact iff property 1 holds iff property 2 holds.

Essentially, Hardy is doing a weaker version of property 1 $\Leftrightarrow$ property 2.

I would also like to point out that his approach is odd, and you would be better reading something like Rudin's principles of Mathematical Analysis if you're learning this stuff for the first time (though sometimes reading Hardy's book is nice for a change).

Answer 2

In a more frequent setting, instead of assuming (2) it is assumed that the closed (hence compact) interval $[a,b]$ is covered by open intervals. Then the claim is that already finitely many of them covers the whole $[a,b]$. As @lyj commented, it is not assumed that $(a,b)$ is in the cover ($\in I$).

For illustration, we can cover $[0,3]$ by $I=\{(-1,2),\ (1,4)\}$, and so on... In your setting, these would be chopped down as $(0,2)$ and $(1,3)$ but that doesn't really make difference: the cover of endpoints are guaranteed by condition (2).

Answer 3

I'll brush some things under the carpet (namely in $\mathbb{R}$ a subset $X$ is compact iff it is sequentially compact) and try to give a statement of Heine-Borel as it would be understood today, and to give some motivation as to why it should be true.

Let $X$ be a set, let $(a_n)$ be a sequence in $X$. When can we say that $(a_n)$ has a convergent subsequence?

For starters, if $X$ is an unbounded set, then we won't really have much of a chance (for example, we can take $(a_n)$ such that $a_n = n$ and this has no convergent subsequence). So we'd better restrict our attention to bounded subsets of $\mathbb{R}$.

Slightly more subtly, we require a more technical condition, which can perhaps be motivated by considering the sequence $a_n = 1/n$ on the set $(0,1)$ (the open interval between 0 and 1 but not containing either endpoint). $a_n \in (0,1)$ for each $n$, but the limit of $a_n$ is 0, and this isn't in $(0,1)$. So we'd better impose the restriction that $X$ should contain all of its limit points (i.e. that $X$ is closed).

It turns out that this is all we need: a set $X$ is sequentially compact (every sequence in $X$ has a convergent subsequence in $X$) if and only if it is closed and bounded. And that's how you'd find Heine-Borel stated today.