I know there are a few posts about the Heine-Borel theorem on the site, however, I have not found a post asking about this specific proof, i.e. from John K. Hunter's lecture notes (Chapter 5, Theorem 5.56).
Theorem 5.56 (Heine-Borel). A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded.
Here is the proof and my questions.
Proof. … First, we prove that a closed, bounded interval $K=[a, b]$ is compact. Suppose that $\mathcal{C}=\left\{A_i: i \in I\right\}$ is an open cover of $[a, b]$, and let
$$
B=\{x \in[a, b]:[a, x] \text { has a finite subcover } \mathcal{S} \subset \mathcal{C}\}.
$$
We claim that $\sup B=b$. The idea of the proof is that any open cover of $[a, x]$ must cover a larger interval since the open set that contains $x$ extends past $x$.
Since $\mathcal{C}$ covers $[a, b]$, there exists a set $A_i \in \mathcal{C}$ with $a \in A_i$, so $[a, a]=\{a\}$ has a subcover consisting of a single set, and $a \in B$. Thus, $B$ is non-empty and bounded from above by $b$, so $c=\sup B \leq b$ exists. Assume for contradiction that $c<b$. Then $[a, c]$ has a finite subcover
$$
\left\{A_{i_1}, A_{i_2}, \ldots, A_{i_n}\right\}
$$
with $c \in A_{i_k}$ for some $1 \leq k \leq n$.
- Why does $[a,c]$ have a finite subcover? Maybe equivalently, how do we know that $c\in B$?
Since $A_{i_k}$ is open and $a \leq c<b$, there exists $\delta>0$ such that $[c, c+\delta) \subset A_{i_k} \cap[a, b]$.
- If $A_{i_k}$ is open, there exists $\delta>0$ such that $(c-\delta,c+\delta)\subset A_{i_k}$. Why $[c, c+\delta) \subset A_{i_k} \cap[a, b]$?
Then $\left\{A_{i_1}, A_{i_2}, \ldots, A_{i_n}\right\}$ is a finite subcover of $[a, x]$ for $c<x<c+\delta$, contradicting the definition of $c$, so sup $B=b$.
- Why is $\left\{A_{i_1}, A_{i_2}, \ldots, A_{i_n}\right\}$ a finite subcover of $[a,x]$ for $c<x<c+\delta$?
Moreover, the following argument shows that, in fact, $b=\max B$.
Since $\mathcal{C}$ covers $[a, b]$, there is an open set $A_{i_0} \in \mathcal{C}$ such that $b \in A_{i_0}$. Then $(b-\delta, b+\delta) \subset A_{i_0}$ for some $\delta>0$, and since $\sup B=b$ there exists $c \in B$ such that $b-\delta<c \leq b$. Let $\left\{A_{i_1}, \ldots, A_{i_n}\right\}$ be a finite subcover of $[a, c]$. Then $\left\{A_{i_0}, A_{i_1}, \ldots, A_{i_n}\right\}$ is a finite subcover of $[a, b]$, which proves that $[a, b]$ is compact.
Now suppose that $K \subset \mathbb{R}$ is a closed, bounded set, and let $\mathcal{C}=\left\{A_i: i \in I\right\}$ be an open cover of $K$. Since $K$ is bounded, $K \subset[a, b]$ for some closed bounded interval $[a, b]$, and, since $K$ is closed, $\mathcal{C}^{\prime}=\mathcal{C} \cup\left\{K^c\right\}$ is an open cover of $[a, b]$. From what we have just proved, $[a, b]$ has a finite subcover that is included in $\mathcal{C}^{\prime}$. Omitting $K^c$ from this subcover, if necessary, we get a finite subcover of $K$ that is included in the original cover $\mathcal{C}$.
To prove the converse, suppose that $K \subset \mathbb{R}$ is compact. Let $A_i=(-i, i)$. Then
$$
\bigcup_{i=1}^{\infty} A_i=\mathbb{R} \supset K
$$
so $\left\{A_i: i \in \mathbb{N}\right\}$ is an open cover of $K$, which has a finite subcover $\left\{A_{i_1}, A_{i_2}, \ldots, A_{i_n}\right\}$. Let $N=\max \left\{i_1, i_2, \ldots, i_n\right\}$. Then
$$
K \subset \bigcup_{k=1}^n A_{i_k}=(-N, N)
$$
so $K$ is bounded.
To prove that $K$ is closed, we prove that $K^c$ is open. Suppose that $x \in K^c$. For $i \in \mathbb{N}$, let
$$
A_i=\left[x-\frac{1}{i}, x+\frac{1}{i}\right]^c=\left(-\infty, x-\frac{1}{i}\right) \cup\left(x+\frac{1}{i}, \infty\right) .
$$
Then $\left\{A_i: i \in \mathbb{N}\right\}$ is an open cover of $K$, since
$$
\bigcup_{i=1}^{\infty} A_i=(-\infty, x) \cup(x, \infty) \supset K
$$
Since $K$ is compact, there is a finite subcover $\left\{A_{i_1}, A_{i_2}, \ldots, A_{i_n}\right\}$. Let $N=$ $\max \left\{i_1, i_2, \ldots, i_n\right\}$. Then
$$
K \subset \bigcup_{k=1}^n A_{i_k}=\left(-\infty, x-\frac{1}{N}\right) \cup\left(x+\frac{1}{N}, \infty\right),
$$
which implies that $(x-1 / N, x+1 / N) \subset K^c$. This proves that $K^c$ is open and $K$ is closed.
- I do not understand how $K \subset \left(-\infty, x-\frac{1}{N}\right) \cup\left(x+\frac{1}{N}, \infty\right)$ implies $(x-1 / N, x+1 / N) \subset K^c$. By $\subset$, the author does not mean proper subset, but rather $\subseteq$ (see page 3 chapter 1):
In our notation, $A \subset X$ does not imply that $A$ is a proper subset of $X$ (that is, a subset of $X$ not equal to $X$ itself), and we may have $A=X$.
Best Answer