Heine-Borel Theorem

Remark

Remark. We say that $f$ is bounded on a set $X$ if there exists a number $M$ such that $|f(x)| < M$ for every $x \in X \cap (\text{ domain } f)$. 
If $f$ is bounded on $X$ and $Y$, then $f$ is bounded on $X \cup Y$. By induction, we see that if $f$ is bounded on $X_1, ..., X_n$, then $f$ is bounded on $X_1 \cup \cdots \cup X_n.$
($\because$) Let $f$ be a funciton from $D \subset \mathbb{R}$ into $\mathbb{R}$. If $f$ is bounded on $X$ and $Y$, then $\exists M_1, M_2 \in \mathbb{R}$ such that $|f(x)| < M_1, \forall x \in X \cap D$ and $|f(x)| < M_2, \forall x \in Y \cap D$. Let $M = \min \{ M_1, M_2 \}$ then $|f(x)| < M, \forall x \in (X \cap D) \cup (Y \cap D) = (X \cup Y) \cap D$. Thus $f$ is bounded on $X \cup Y$. 

Lemma 34.1

Lemma 34.1. If \( f \) is continuous at \( c \), there exists \( \delta > 0 \) such that \( f \) is bounded on \( (c - \delta, c + \delta) \).

Proof. By Theorem 33.3, $\forall \varepsilon > 0, \exists \delta > 0$ such that if $x$ is in the domain of $f$ and $|x - c| < \delta$, then $|f(x) - f(c)| < \varepsilon$. This means that $$x \in (c - \delta, c + \delta) \\ \Longrightarrow |f(x)| = |f(x) - f(c) + f(c)| \leq |f(x) - f(c)| + |f(c)| < \varepsilon + |f(c)|.$$ Thus $f$ is boundd on $(c - \delta, c+ \delta)$. $\blacksquare$

Heine–Borel Theorem

Theorem 34.2 (Heine–Borel Theorem). Let \( \mathcal{J} \) be a collection of open intervals such that \[ \bigcup \mathcal{J} \supset [a, b] \] Then there exists a finite subset \( \{ I_1, \dots, I_n \} \) of \( \mathcal{J} \) such that \[ \bigcup_{i=1}^n I_i \supset [a, b]. \]

Proof. Let $$X = \bigg \{ x \in (a,b] \, | \, [a, x] \subset \bigcup_{i=1}^n I_i \text{ for some } I_1, ..., I_n \in \mathcal{J} \bigg \}.$$ Since $a \in \cup \mathcal{J}$, $a \in I = (s_0, t_0)$ for some $I \in \mathcal{J}$. If $b \in I$, then the proof is done. Assume that $b \notin I$, so that $t_0 \leq b$. Then we can choose $x_0$ such that $a < x_0 < t_0$, so that $x_0 \in X$. Since $X$ is bounded above by $b$, $X$ has the least upper bound $c$ by A13. Note that $a < x_0 \leq c \Longrightarrow a < c$. Since $b$ is an upper bound for $X$, $c \leq b$, so that $c \in (a, b]$ and $c \in \cup \mathcal{J}$.
Then $c \in I' = (s_1, t_1)$ for some $I' \in \mathcal{J}$. Since $s_1 < c$, $s_1$ is not an upper bound for $X$, which means that $\exists x \in X$ such that $s_1 < x \leq c$. By the definition of $X$, there exists $I_1, ..., I_n \in \mathcal{J}$ such that $$[a, x] \subset \bigcup_{i=1}^n I_i \\ \Longrightarrow [a, c] \subset \left[ \bigcup_{i=1}^n I_i \right] \cup (s_1, t_1).$$ This menas that $c \in X$. 
Suppose that $c < b$. Since $c \in X$, there exists $I_1, ..., I_n \in \mathcal{J}$ such that $$[a, c] \subset \bigcup_{i=1}^n I_i.$$ Then $c \in I_j = (s_2, t_2)$ for some $j (1 \leq j \leq n)$. Note that $c < b$ and $c < t_2$. Choose a number $d$ such that $c < d < \min \{ b, t_2 \}$. Then $$[a, d] \subset \bigcup_{i=1}^n I_i \\ \Longrightarrow d \in X \\ \Longrightarrow d \leq c \quad ((\because) c \text{ is an upper bound for } X.) \bigotimes$$ Hence $c = b \in X$, which means that there exists $I_1, ... I_n \in \mathcal{J}$ such that $$[a, b] \subset \bigcup_{i=1}^n I_i. \blacksquare$$

Theorem 34.3

Theorem 34.3. If \( f \) is continuous on \( [a, b] \), then \( f \) is bounded on \( [a, b] \).

Proof. By Lemma 34.1, for each $c \in [a, b]$, $f$ is bounded on an open interval $I_c$. Note that $$[a, b] \subset \bigcup_{c \in [a, b]} I_c.$$ By Heine-Borel Theorem, there exists $c_1, ..., c_n \in [a, b]$ such that $[a, b] \subset \bigcup_{i=1}^n I_{c_i}$. Thus $f$ is bounded on $\bigcup_{i=1}^n I_{c_i}$, so that $f$ is bounded on $[a, b]$. $\blacksquare$ 

Theorem 34.4

Theorem 34.4. If \( f \) is continuous on \( [a, b] \), there exist points \( c \) and \( d \) in \( [a, b] \) such that \[ f(c) \leq f(x) \leq f(d) \] for all \( x \in [a, b] \). That is, if \( f \) is continuous on \( [a, b] \), then \( f \) attains a maximum and a minimum on \( [a, b] \).

Proof. By Theorem 34.3, $f$ is bounded on $[a, b]$. Then there is the least upper bound $M = \sup \{ f(x) \, | \, x \in [a, b] \}$. 
Suppose that $f(x) \neq M, \forall x \in [a, b]$. Then $f(x) < M, \forall x \in [a, b]$. Let $$g(x) = \frac{1}{M - f(x)}, \quad x \in [a, b].$$ Since $g$ is continuous on $[a, b]$ by Theorem 33.2, $g$ is bounded on $[a, b]$ by Theorem 34.3, so that $\exists N$ such that $|g(x)| < N, \forall x \in [a, b]$. Then for all $x \in [a, b]$, we have that $$g(x) = \frac{1}{M - f(x)} < N \\ \Longrightarrow f(x) < M - \frac{1}{N} < M.$$ But $M - \frac{1}{N}$ is an upper bound for $\{ f(x) \, | \, x \in [a, b] \}$, so that $M \leq M - \frac{1}{N}.$ $\bigotimes$ Thus $\exists d \in [a, b]$ such that $f(d) = M$, so that $f(x) \leq f(d), \forall x \in [a, b]$. 
Similarly, we can show that $\exists c \in [a, b]$ such that $f(c) \leq f(x), \forall x \in [a, b]$. $\blacksquare$