108學年度｜微積分模組班｜Class 01｜Note 2.

Note 2. Completeness of $\mathbb{R}$ & IVT / IFT

• Date: 108/10/14 (Week 06)
• Place: 新生教學館 301

Part 1: Completeness of Real number $\mathbb{R}$

There are various way to construct real number system. Depending on the construction, we may regard some of the statement in Proposition 1.1 as axioms$^1$. This relate to some tedious theory and mathematical proof (though, essential), which is not what we are discussing here.

($^1$: Axioms (公設/公理)：指的是數學上定義性的命題，常常用於建構特定數學領域(算數、代數、分析等)。比方說，在代數上我們定義一個原群(Magma)是一個有二元運算的集合(可以先把二元運算理解為像是加法的東西)。在這個例子中，「有二元運算」就是原群的公設。)

In short, we will not "prove" if any statement in Proposition 1.1 holds in $\mathbb{R}$, but showing they are logically equivalent. How will this even benefit us? Well, if the equivalence is proved, then by admitting one of the statement, we will immediately have the other three property.

Proposition 1.1

• The following are equivalent.

1. Completeness.
2. Monotone convergence theorem.
3. Least upper bound property.
4. Nested interval theorem.

We will clarify the terms here and prove this proposition in the last of this section.

1.1 Completeness

Definition 1.1 

• A metric (距離函數) on a set $S$ is a function $d \colon S \times S \to \mathbb{R}_{\geqslant 0}$ satisfying:

1. For all $x,y \in S$, we have $d(x,y) = 0 \Longleftrightarrow x=y$ (Identity of indiscernibles)
2. For all $x,y \in S$, we have $d(x,y) = d(y,x)$ (Symmetry)
3. For all $x,y,z \in S$, we have $d(x,z) \leqslant d(x,y) + d(y,z)$ (Trianglular inequality)

Remark 1.2. 

• In particular, a set together with a metric $(S,d)$ is called a metric space (度量空間)。

Example 1.3.

1. The real number $\mathbb{R}$ with standard metric $d(x,y) = |x-y|$ forms a metric space.
2. The $n-$dimensional Euclidean space $\mathbb{R}^n$ with standard metric: $$d(\textbf{x},\textbf{y}) = \sqrt{\sum_{i = 1}^n (x_i-y_i)^2}$$ where $\textbf{x} = (x_1,x_2, \cdots, x_n)$, $\textbf{y} = (y_1,y_2, \cdots, y_n)$ forms a metric space.
3. A decrete metric on a set $S$ is defined by: $$d(x,y) = \begin{cases} 1 &, \text{ if } x \neq y \\ 0 &, \text{ if } x = y \end{cases}$$

Definition 1.4.

1. In a metric space $(S,d)$, we say a sequence $\{a_n\}_{n = 1}^\infty$ converges to $L$ if for any $\epsilon > 0$, there exist $n_0 \in \mathbb{N}$ such that $d(x_n,L) < \epsilon$. If this is the case, we say $\{a_n\}_{n = 1}^\infty$ is convergent and denote by: $$\lim_{n \to \infty} a_n = L$$
2. In a metric space $(S,d)$, we say a sequence $\{a_n\}_{n = 1}^\infty$ is Cauchy if for any $\epsilon > 0$, there exist $n_0 \in \mathbb{N}$ such that $d(x_n,x_m) < \epsilon$ when $n,m \geqslant n_0$.

Example 1.5.

1. In $\mathbb{R}$ together with the standard metric, $\left\{a_n = \dfrac{1}{n} \right\}_{n = 1}^\infty$ is both a convergent and Cauchy sequence.
2. In $\mathbb{Q}$ together with the standard metric, $\left\{a_n = \left(1 + \dfrac{1}{n}\right)^n \right\}_{n = 1}^\infty$ is Cauchy but not convergent. (Since the Euler number $e \not \in \mathbb{Q}$)

Definition 1.6.

• We say a metric space $(S,d)$ is complete if every Cauchy sequence converges.

Theorem 1.7. (Completeness of $\mathbb{R}$)

• $\mathbb{R}$ is complete under the standard metric $d(x,y) = |x-y|$.

1.2 Monotone Convergence Theorem

Definition 1.8.

1. A sequence $\{a_n\}_{n = 1}^\infty \subset \mathbb{R}$ is increasing if $a_n \leqslant a_{n+1}$ for all $n \in \mathbb{N}$.
2. A sequence $\{a_n\}_{n = 1}^\infty \subset \mathbb{R}$ is decreasing if $a_n \geqslant a_{n+1}$ for all $n \in \mathbb{N}$.

Theorem 1.9. (Monotone Convergence Theorem)

• An increasing sequence in $\mathbb{R}$ which is bounded above must converge.

Exercise 1.10.

1. Show that the following are equivalent:
1. An increasing sequence in $\mathbb{R}$ which is bounded above must converge.
2. A decreasing sequence in $\mathbb{R}$ which is bounded below must converge.
2. Let $a_1 = \sqrt{2}$ and define $a_{n + 1} = \sqrt{2 + a_n}$ for all $n \in \mathbb{N}$, show that:
1. $\{a_n\}_{n = 1}^\infty$ is an increasing sequence. (Hint: induction)
2. $\{a_n\}_{n = 1}^\infty$ is bounded above. (Hint: induction)
3. $\{a_n\}_{n = 1}^\infty$ is convergent and find its limit.

1.3 Least Upper Bound Property

Definition 1.11.

1. The least upper bound of a set $S \subset \mathbb{R}$ is an upper bound that doesn't exceed any other upper bound, usually denote by $\sup S$, the supremum of $S$.
2. The greatest lower bound of a set $S \subset \mathbb{R}$ is a lower bound that exceeds any other upper bound, usually denote by $\inf S$, the infimum of $S$.

Example 1.12.

1. $\sup [0,1] = \sup (0,1) =\sup (-\infty, 1) = 1$.
2. Supremum of $(0,\infty)$ does not exist, infimum of $\mathbb{Q}$ does not exist.

Theorem 1.13. (Least upper bound property)

• Every subset $S \subset \mathbb{R}$ bounded above has a least upper bound.

1.4 Nested Interval Theorem

Theorem 1.14. (Nested interval theorem)

• Let $\{I_n = [a_n, b_n]\}_{n = 1}^\infty$ be a sequence of intervals in $\mathbb{R}$ satisfying:
1. $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$. (The interval is "nested")
2. $\displaystyle \lim_{n \to \infty}(b_n-a_n)=0$
Then we have:

1. The sequence $\{a_n\}_{n = 1}^\infty$ and $\{b_n\}_{n = 1}^\infty$ converges and $\displaystyle \lim_{n \to \infty} a_n =\lim_{n \to \infty} b_n$.
2. The intersection $\displaystyle \bigcap_{n = 1}^\infty I_n$ is not empty, moreover, it contains only one element which is the limit of both $a_n$ and $b_n$.

Example 1.15.

1. Let $I_n = \left[ 0,\dfrac{1}{n}\right]$, obviously, $I_{n+1} \subset I_{n}$ and $\displaystyle \lim_{n \to \infty} \left(\dfrac{1}{n} - 0\right) = 0$, so $\displaystyle \bigcap_{n=1}^\infty  \left[ 0,\dfrac{1}{n}\right] \neq \varnothing$. Precisely, $\displaystyle \bigcap_{n=1}^\infty  \left[ 0,\dfrac{1}{n}\right] = \{0\}$, where $0$ is the limit of the left end and right end of the sequence of intervals.
2. What is the importance of requiring "closed interval" in Theorem 1.14? Take \[I_n = \left(0,\dfrac{1}{n}\right)\], verify that it does satisfy the condition 1.,2. in \textbf{Theorem 1.14}, but $\displaystyle \bigcap_{n = 1}^\infty \left(0,\dfrac{1}{n}\right) = \varnothing$.

1.5 Conclusion

Finally, we have enough knowledge and familiarity of mathematical terms to prove the following which is stated in the previous paragraph:

Proposition 1.1

• The following are equivalent.

1. Completeness.
2. Monotone convergence theorem.
3. Least upper bound property.
4. Nested interval theorem.

Proof.

• The strategy of the proof is to show the implication: $$1. \Longrightarrow 2. \Longrightarrow 3. \Longrightarrow 4. \Longrightarrow 1.$$
• (1. Completeness of $\mathbb{R} \Longrightarrow$ 2. Monotone convergence theorem)
Let $\{a_n\}_{n = 1}^\infty$ be an increasing sequence in $\mathbb{R}$ which is bounded above, we ought to prove that it is convergent. We claim the following:
1. Claim: $\{a_n\}_{n = 1}^\infty$ is Cauchy.
Suppose not, that is, there exists $\epsilon_0 > 0$ such that for every $N \in \mathbb{N}$, there is $n > N$ such that $a_n - a_N \overset{(*)}{=} |a_n-a_N| > \epsilon_0$. $(*)$ holds since $\{a_n\}_{n = 1}^\infty$ is increasing. Pick $n_1 = 1$, there is $n_2 > n_1$ such that $a_{n_2}-a_{n_1} = |a_{n_2}-a_{n_1}|>\epsilon_0$. Again, there exist $n_3 > n_2$ such that $a_{n_3}-a_{n_2} = |a_{n_3}-a_{n_2}|>\epsilon_0$, and so on. By such, we may construct a subsequence $\{a_{n_k}\}_{k = 1}^\infty$ of $\{a_n\}_{n = 1}^\infty$ which satisfy that for all $k \in \mathbb{N}$. $$a_{n_{k+1}} - a_{n_k} > \epsilon_0$$ Now suppose $B$ is an upper bound of $\{a_n\}_{n = 1}^\infty$, let $K = \left \lceil \dfrac{B-a_1}{\epsilon_0}\right \rceil$, then: $$a_{n_K} > a_{n_1} + \left \lceil \dfrac{B-a_1}{\epsilon_0}\right \rceil \cdot \epsilon_0 \geqslant a_1 + (B-a_1) = B $$, a contradiction to the fact that $\{a_n\}_{n = 1}^\infty$ is bounded, so $\{a_n\}_{n = 1}^\infty$ must be Cauchy.

2. Apply our assumption (3. Completeness of $\mathbb{R}$), we know Cauchy sequence converges, hence $\{a_n\}_{n = 1}^\infty$ converges, which proves the Monotone convergence theorem.
• (2. Monotone convergence theorem $ \Longrightarrow$ 3. Least upper bound property)
Let $S \subset \mathbb{R}$ be a bounded set, and let $U = \{\text{upper bounds of } S\}$, we ought to prove that there is a smallest element in $U$. Define the following sequence: \begin{align*} B_1 &= \min\left\{ \dfrac{k}{2} \in U \mid k \in \mathbb{Z}\right\} \\ B_1 &= \min\left\{ \dfrac{k}{2^2} \in U \mid k \in \mathbb{Z}\right\} \\ \vdots & \hspace{3cm} \vdots \\ B_n &= \min\left\{ \dfrac{k}{2^n} \in U \mid k \in \mathbb{Z}\right\} \end{align*}
1. Claim : $\{B_n\}_{n=1}^{\infty}$ is decreasing and bounded below.
Suppose $\{B_n\}_{n=1}^{\infty}$ is not decreasing, which means there is $n \in \mathbb{N}$ such that $B_{n+1}>B_n$. Let $B_n=\dfrac{k_n}{2^n}$, $B_{n+1}=\dfrac{k_{n+1}}{2^{n+1}}$ for some $k_n,k_{n+1} \in \mathbb{Z}$, then : \begin{align*} B_{n+1}>B_n &\Longrightarrow \frac{k_{n+1}}{2^{n+1}}-\frac{k_n}{2^n} >0 \\ &\Longrightarrow k_{n+1}-2k_n>0 \\ &\Longrightarrow k_{n+1}-2k_n \geqslant 1 \\ &\Longrightarrow \frac{k_{n+1}}{2^{n+1}}-\frac{k_n}{2^n} \geqslant \frac{1}{2^{n+1}} \\ &\Longrightarrow B_{n+1}-\frac{1}{2^{n+1}} \geqslant B_n \end{align*} Since $B_n$ is an upper bound, so $B_{n+1}-\dfrac{1}{2^{n+1}}$ is an upper bound, but this contradicts with our construction of $B_n$ (Why?), hence $B_{n+1} \leqslant B_n$. And obviously, $\{B_n\}_{n=1}^{\infty}$ is bounded below. (Why?) Hence $\{B_n\}_{n=1}^{\infty}$ is decreasing and bounded below, where by our assumption (2. Monotone convergence theorem), $\displaystyle \lim_{n \to \infty}B_n$ exists, denote $\displaystyle B=\lim_{n \to \infty}B_n$.

2. Claim : $B = \sup S$.
Notice $s \leqslant B_n$ for all $s \in S$, hence: $$ s \leqslant B=\lim_{n \to \infty}B_n$$, giving $B$ is an upper bound of $S$. Also, by our construction, $\displaystyle B_n-\frac{1}{2^n}$ is not an upper bound, meaning that there exists $s_n \in S$ such that $\displaystyle s_n \geqslant B_n-\frac{1}{2^n}$, $\forall n \in \mathbb{N}$, Let $u \in U$ be an upper bound of $S$, then $\displaystyle B_n-\frac{1}{2^n} \leqslant s_n \leqslant u$, $\forall n \in \mathbb{N}$. Hence $\displaystyle B=\lim_{n \to \infty}\left(B_n-\frac{1}{2^n}\right) \leqslant u \Rightarrow B$ does not exceed any upper bound, so $B=$ sup $S$.

3. We may thus conclude the existence of the supremum of $S$.
• (3. Least upper bound property $\Longrightarrow$ 4. Nested interval theorem)
Let $\{I_n = [a_n, b_n]\}_{n = 1}^\infty$ be a sequence of intervals in $\mathbb{R}$ satisfying:
1. $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$.
2. $\displaystyle \lim_{n \to \infty}(b_n-a_n)=0$
, we ought to prove the following:
1. The sequence $\{a_n\}_{n = 1}^\infty$ and $\{b_n\}_{n = 1}^\infty$ converges and $\displaystyle \lim_{n \to \infty} a_n =\lim_{n \to \infty} b_n$.
2. The intesection $\displaystyle \bigcap_{n = 1}^\infty I_n$ is not empty, moreover, it contains only one element which is the limit of both $a_n$ and $b_n$.
Note that $\{a_n\}_{n =1}^\infty$ has an upper bound $b_1$, similarly, $\{b_n\}_{n =1}^\infty$ ha a lower bound $a_1$, by assumption (3. Least upper bound property), there exist supremum of $\{a_n\}_{n =1}^\infty$, say, $A$; and there exist infimum of $\{b_n\}_{n =1}^\infty$, say, $B$.
1. Claim: $\displaystyle \lim_{n \to \infty}a_n = A$ and $\displaystyle \lim_{n \to \infty} b_n = B$, moreover, $A=B$.
Note that for every $k \in \mathbb{N}$, $A-\dfrac{1}{k}$ can't be an upper bound of $\{a_n\}_{n =1}^\infty$, so there exist $a_{n_k}$ such that $a_{n_k} > A-\dfrac{1}{k}$ for all $k \in \mathbb{N}$. Precisely, we have: $$A-\dfrac{1}{k} < a_{n_k} \leqslant A$$, for all $k \in \mathbb{N}$, so $\displaystyle \lim_{k \to \infty}a_{n_k} = A$, since $\{a_n\}_{n =1}^\infty$ is increasing, so $\displaystyle \lim_{n \to \infty}a_{n} = A$. (Why?) Similarly, $\displaystyle \lim_{n \to \infty}b_n = B$. Last, since we require $\displaystyle \lim_{n \to \infty}(b_n-a_n)=0$, so $A=B$, we denote $L = A = B$ here:
2. Claim: $\displaystyle \bigcap_{n = 1}^\infty I_n = \{L\}$.
Note that $a_n \leqslant L \leqslant b_n$ for all $n \in \mathbb{N}$, so $\displaystyle \{L\} \subset \bigcap_{n = 1}^\infty I_n$. To prove the other direction of inclusion, suppose there is another $\displaystyle L \neq M \in \bigcap_{n = 1}^\infty I_n$. WLOG, suppose $L<M$, then $\displaystyle [L,M] \subset \bigcap_{n = 1}^\infty I_n$ (Why?) a contradiction, since this implies $$0<M-L \leqslant \lim_{n \to \infty} (b_n-a_n) = 0$$, so $\displaystyle \{L\} \supset \bigcap_{n = 1}^\infty I_n$, and hence $\displaystyle \{L\} = \bigcap_{n = 1}^\infty I_n$. By, part 1., 2., we proved the Nested interval theorem.
• (4. Nested interval theorem $ \Longrightarrow$ 1. Completeness of $\mathbb{R}$
Let $\{a_n\}_{n = 1}^\infty$ be a Cauchy sequence in $\mathbb{R}$, we ought to prove that it is convergent. Note that for all $k \in \mathbb{N}$, there exist $n_k \in \mathbb{N}$ such that: $$|a_n-a_m|<\dfrac{1}{k}, \text{ for all } n,m \geqslant n_k$$, since $\{a_n\}_{n = 1}^\infty$ is Cauchy. We define the following sequence of intervals: \begin{align*} I_1 &= \left[a_{n_1} - \dfrac{1}{1}, a_{n_1} + \dfrac{1}{1}\right] \\ I_2 &= \left[a_{n_2} - \dfrac{1}{2}, a_{n_2} + \dfrac{1}{2}\right] \cap I_1 \\ \vdots & \hspace{3cm} \vdots \\ I_k &= \left[a_{n_k} - \dfrac{1}{k}, a_{n_k} + \dfrac{1}{k}\right] \cap I_{k-1} \end{align*}
, then obviously $I_{k+1} \subset I_k$ for all $k \in \mathbb{N}$ and $\displaystyle \lim_{k \to \infty}|I_k| \leqslant \lim_{k \to \infty}\dfrac{2}{k} = 0$. By assumption (4. Nested interval theorem), there exist a unique number $L \in \mathbb{R}$ such that: $$\bigcap_{n = 1}^\infty I_n = \{L\}$$
Claim: $\displaystyle \lim_{n \to \infty}a_n = L$.
For any $\epsilon > 0$, there is a $K \in \mathbb{N}$ such that $\dfrac{2}{K} < \epsilon$, hence for all $n \geqslant n_K$, we have: $$|a_n - L| \leqslant |a_n - a_{n_K}| + |a_{n_K}-L| < \dfrac{1}{K} + \dfrac{1}{K} < \epsilon$$
, so $\displaystyle \lim_{n \to \infty}a_n = L$.
And thus, we prove the completeness of $\mathbb{R}$.
• Summarize of what we just proved: $$\begin{tikzcd} 1. Completeness of $\mathbb{R}$ \arrow{r}{1^\circ} & \textbf{2. Monotone convergence theorem} \arrow{d}{2^\circ} \\ \textbf{4. Nested interval theorem} \arrow{u}{4^\circ} & \textbf{3. Least upper bound property} \arrow{l}{3^\circ} \end{tikzcd}$$

$\square$

Part 2: Intermediate Value Theorem

Theorem 2.1. (Intermediate Value Theorem)

• Suppose $f \colon [a,b] \to \mathbb{R}$ is continuous on $[a,b]$, then for any $y_0 \in (f(a),f(b))$ (WLOG, we suppose $f(b)>f(a)$ here), there exist $x_0 \in (a,b)$ such that $f(x_0) = y_0$.

Proof.

1. Define $S = \{x \in [a,b] \mid f(x) \leqslant y_0\}$, this set is nonempty since $a \in S$. Applying least upper bound property (\textbf{Theorem 1.13}), there exist $c = \sup S$. We claim that $c \in (a,b)$ and $f(c) = y_0$.
2. (Proving $c \in (a,b)$)
Note that $S \subset [a,b]$, so $c = \sup S \leqslant \sup [a,b] = b$, and obviously we have $c \geqslant a$, so $c \in [a,b]$. To show stonger that $c \in (a,b)$, it relies on the after fact that: $$f(c) = y_0 \in (f(a),f(b))$$ , hence $c \neq a$, $c \neq b$, which is, $c \in (a,b)$.

3. (Proving $f(c) = y_0$)
Suppose not, that is, $f(c)>y_0$ or $f(c)<y_0$, we will discuss in the following.
1. Suppose $f(c) > y_0$.
Let $\epsilon = f(c) - y_0>0$, since $f(x)$ is continuous, there exist $\delta >0$ such that: $$|f(x)-f(c)|<\epsilon \text{ for all }|x-c|<\delta$$, that is, for all $|x-c|<\delta$, we have: \begin{equation} y_0 < f(x) < 2f(c) - y_0 \end{equation}, which is impossible, since $c = \sup S$, there exist $x_0 \in (c-\delta, c)$ such that $x_0 \in S$, that is, $f(x_0) \leqslant y_0$, a contradiction to inequality $(1)$.
2. Suppose $f(c) < y_0$.
Let $\epsilon = y_0 - f(c)>0$, since $f(x)$ is continuous, there exist $\delta >0$ such that: $$|f(x)-f(c)|<\epsilon \text{ for all }|x-c|<\delta$$, that is, for all $|x-c|<\delta$, we have: \begin{equation} 2f(c) - y_0 < f(x) < y_0 \end{equation}, which is impossible, since $c = \sup S$, so for all $x \in (c, c+\delta)$, we have $x \not \in S$, that is, $f(x_0) > y_0$, a contradiction to inequality $(2)$.
Hence, $f(c) = y_0$.

$\square$

Part 3: Inverse Function Theorem (1-dimension-case)

Theorem 3.1. (Inverse Function Theorem (1-dimension-case))

• Let $f(x)$ be a strictly increasing, continuous function on $(a,b)$. Then we have:

1. There exists an inverse function $g(y) \colon (f(a), f(b)) \to (a,b)$.
2. $g(y)$ is continuous on $(f(a), f(b))$.
3. Suppose furthermore that $f(x)$ is differentiable on $(a,b)$, then $g(y)$ is differentiable on $(f(a), f(b))$ and $g'(y_0) = \dfrac{1}{f'(x_0)}$ if $y_0 = f(x_0)$.

Proof.

1. Since $f(x)$ is strictly increasing, it is bijective and hence possess an inverse function $g(y) \colon (f(a), f(b)) \to (a,b)$.
2. Pick $y_0 \in (f(a),f(b))$, suppose $f(x_0)=y_0$ with $a < x_0 < b$, pick $$0< \epsilon < \min\{ x_0-a, b-x_0 \}$$ (The restriction of $\epsilon$ is only to ensure that such $\epsilon$ doesn't give a interval larger than $(a,b)$, it actually doesn't matter if we have a larger $\epsilon >0$) For this $\epsilon$, we have: $$a<x_0-\epsilon<x_0<x_0+\epsilon<b$$, and since $f(x)$ is strictly increasing, we have: $$f(a)<f(x_0-\epsilon)<f(x_0)<f(x_0+\epsilon)<f(b)$$ Let $\delta=\min\{f(x_0)-f(x_0-\epsilon), f(x_0+\epsilon)-f(x_0)\}$. If $|y-y_0|=|y-f(x_0)|<\delta$, we have:\begin{align*} f(x_0-\epsilon)<y<f(x_0+\epsilon) & \overset{(*)}{\Longleftrightarrow} g(f(x_0-\epsilon))<g(y)<g(f(x_0+\epsilon)) \\ & \Longleftrightarrow  x_0-\epsilon<g(y)<x_0+\epsilon \\ & \Longleftrightarrow |g(y)-x_0|<\epsilon \\ & \Longleftrightarrow |g(y)-g(y_0)|<\epsilon \end{align*}, where $(*)$ holds since $g(y)$ is also strictly increasing. (Why?) Hence, $g(y)$ must be continuous on $[f(a),f(b)]$.
3. Fix $y_0 \in (f(a), f(b))$ with $f(x_0) = y_0$. Note that $f(x)$ is differentiable at $x = x_0$, that is, $$\lim_{x \to x_0}\dfrac{f(x)-f(x_0)}{x-x_0} = f'(x_0).$$ Since $f(x)$ is strictly increasing, so $f'(x_0) \neq 0$ and $f(x)-f(x_0) \neq 0$ whenever $x \neq x_0$, hence we have: $$\lim_{x \to x_0} \dfrac{x-x_0}{f(x)-f(x_0)} = \lim_{x \to x_0} \dfrac{1}{\dfrac{f(x)-f(x_0)}{x-x_0}} = \dfrac{1}{f'(x_0)}$$ That is, for any $\epsilon>0$, there is $\delta >0$ such that: $$\left|\dfrac{x-x_0}{f(x)-f(x_0)} - \dfrac{1}{f'(x_0)}\right|<\epsilon$ when $|x-x_0|<\delta$$ Since $g(y)$ is continuous, there exist $\gamma >0$ such that $|x-x_0| = |g(y) - g(y_0)|<\delta$ when $|y-y_0|<\gamma$. In conclusion, when $|y-y_0|<\gamma$, we have: $$\left|\dfrac{g(y)-g(y_0)}{y-y_0} - \dfrac{1}{f'(x_0)}\right| = \left|\dfrac{x-x_0}{f(x)-f(x_0)} - \dfrac{1}{f'(x_0)}\right| < \epsilon$$ That is precisely $\displaystyle \lim_{y \to y_0}\dfrac{g(y)-g(y_0)}{y-y_0} = \dfrac{1}{f'(x_0)}$, so $g(y)$ is differentiable at $y = y_0$ and hence on $(f(a),f(b))$.  
    $\square$