I am studying complex analysis from the second edition of the book Foundations of Complex Analysis written by S.Ponnusamy. While going through the section $4.6$ (Homotopy Version of Cauchy's Theorem) of this book in page no. $146,$ I came across a definition $4.51$ in page no. $147$ which states as follows $:$
"Let $D$ be a domain (open and connected set) in $\Bbb C$ and $\gamma_0$ and $\gamma_1$ be two curves defined on $[0,1].$ We say that $\gamma_0$ and $\gamma_1$ are close together if there exists a partition $\mathcal P$ of $[0,1],$ $\mathcal P : 0 = t_0 < t_1 < \cdots < t_n = 1,$ and a sequence of disks $D_j,$ $j=0,1,\cdots, n-1,$ such that for each $j=0,1,\cdots,n-1,$ $D_j$ contains the images $\gamma_0 \left ([t_k,t_{k+1}] \right )$ and $\gamma_1 \left ([t_k,t_{k+1}] \right ).$"
With this definition in mind the author proved the following theorem (Theorem $4.52$ in his book in the page no. $148$) $:$
Theorem $:$ Let $D$ be a domain in $\Bbb C$ and $\gamma_0$ and $\gamma_1$ be two closed contours in $D.$ Suppose that $\gamma_0$ and $\gamma_1$ are close together. Then, for each $f \in \mathcal H(D),$ we have $$\int_{\gamma_0} f(z)\ dz = \int_{\gamma_1} f(z)\ dz.$$
With the help of the above theorem the author proved Homotopy Version of Cauchy's Theorem in the page no. $149$ (Theorem $4.54$). Here it is $:$
Theorem $:$ Let $D$ be a domain in $\Bbb C$ and $\gamma_0$ and $\gamma_1$ be two closed contours in $D$ such that $\gamma_0 \simeq \gamma_1$ (Fixed Endpoint Homotopic) in $D.$ Then for each $f \in \mathcal H(D),$ we have $$\int_{\gamma_0} f(z)\ dz = \int_{\gamma_1} f(z)\ dz.$$
The proof of the above theorem goes as follows $:$
Proof. Let $F : \gamma_0 \simeq \gamma_1$ be a homotopy in $D.$ Since $F$ is continuous on the square $R = [0,1] \times [0,1]$ which is compact, the image $F(R)$ is compact and $F$ is uniformly continuous on $R.$ Hence $F(R)$ has a positive distance from $\Bbb C \setminus D.$ Choose partitions $$0=u_0<u_1<\cdots<u_m =1\ \ \ \ \ \ \ \ 0=t_0<t_1<\cdots<t_n =1$$ of $[0,1]$ and let $R_{jk} = [t_j,t_{j+1}] \times [u_j,u_{j+1}]$ $(j=0,1,\cdots,m-1;\ k = 0,1,\cdots, n-1),$ a rectangle. Then $F(R_{jk}) = D_{jk} \subseteq D.$ Define $\Gamma_k$ by $$\Gamma_k (t) = F(t,u_k),\ t \in [0,1],\ k=0,1,\cdots,m.$$ Then $\Gamma_k$'s are continuous and the curves $\Gamma_k$ and $\Gamma_{k+1}$ are close togther. Hence by the previuos theorem it follows that $$\int_{\Gamma_k} f(z)\ dz = \int_{\Gamma_{k+1}} f(z)\ dz,\ k=0,1,\cdots,m-1.$$ As $\Gamma_0 = \gamma_0$ and $\Gamma_m = \gamma_1,$ the desired equality follows.
In the above proof I don't understand why $\Gamma_k$ and $\Gamma_{k+1}$ are close together for $k=0,1,\cdots,m-1.$ Can anybody please help me in this regard?
Thanks in advance.
EDIT $:$ Since for $k=0,1,\cdots,m$ the image of $\Gamma_k$ is in $D$ and $D$ is open so for any point $z$ on $\left (\Gamma_k + \Gamma_{k+1} \right )$ there exists $\epsilon_z \gt 0$ such that $B(z,\epsilon_z) \subseteq D.$ Now consider the collection $$\mathcal U: =\left \{B \left (z, \frac {\epsilon_z} {2} \right )\ \bigg |\ z \in \text {Range} \left (\Gamma_k + \Gamma_{k+1} \right) \right \}.$$ Since $\left ( \Gamma_k + \Gamma_{k+1} \right )$ is continuous on the compact set $[0,1]$ the image of $\left (\Gamma_k + \Gamma_{k+1} \right )$ is compact. As $\mathcal U$ is an open cover of the image of $\left (\Gamma_k + \Gamma_{k+1} \right )$ and the image of $\Gamma_k$ is compact, $\mathcal U$ has finite subcover say $\{B(z_r,\epsilon_r)\ |\ r = 1,2, \cdots, l \}.$ Let $\epsilon : = \min \{\epsilon_1,\epsilon_2,\cdots, \epsilon_l \}.$ Since both $\Gamma_k$ and $\Gamma_{k+1}$ are continuous on the compact set $[0,1]$ they arew uniformly continuous and hence there exists $\delta > 0$ such that for any pair of points $t,t' \in [0,1]$ with $|t-t'| \lt \delta$ we have $$\left |\Gamma_k (t) – \Gamma_k (t') \right | < \frac {\epsilon} {2} \ \ \ \ \text{and}\ \ \ \ \left |\Gamma_{k+1} (t) – \Gamma_{k+1} (t') \right | \lt \frac {\epsilon} {2}.$$ Now if we consider the partition $\mathcal P : 0=t_0<t_1<\cdots<t_n = 1$ in such a way that the norm of the partition $\mathcal P$ i.e. $\left \| P \right \| \lt \delta,$ then we are through. Am I right? Please verify my argument.
Note $:$ Here by $\left (\Gamma_k + \Gamma_{k+1} \right )$ I mean the concatenation of $\Gamma_k$ and $\Gamma_{k+1}.$
I think here the main problem is that $\Gamma_k \left ([t_k,t_{k+1}] \right )$ and $\Gamma_{k+1} \left ([t_k,t_{k+1}] \right )$ may not be contained in the same disk in the finite subcover. But the definition of closing together requires that extra condition. How do I get rid of this draw back? Can anybody give me some suggestion?
RE-EDIT $:$ Since $F$ is a homotopy defined on the compact set $R = [0,1] \times [0,1],$ $F(R)$ is compact. Since $F(R) \subseteq D$ so $F(R)$ has some positive distance from $\Bbb C \setminus D.$ Let $\text {dist} (F(R), \Bbb C \setminus D) = \epsilon > 0.$ Then for any $z \in F(R)$ we have $B \left (z, \frac {\epsilon} {2} \right ) \subseteq D.$ Consider the collection $$\mathcal U: = \left \{B \left (z, \frac {\epsilon} {6} \right )\ \bigg |\ z \in F(R) \right \}.$$ Then clearly $\mathcal U$ is an open cover of $F(R).$ Since $F(R)$ is compact $\mathcal U$ has a finite subcover. Let it be $\mathcal U',$ where $$\mathcal U' : = \left \{ B \left (z_r, \frac {\epsilon} {6} \right )\ \bigg |\ r = 1,2, \cdots, p \right \}.$$ Now let us consider the collection of disks $$\mathcal D : = \left \{ B \left (z_r, \frac {\epsilon} {2} \right )\ \bigg |\ r = 1,2, \cdots, p \right \}.$$ Since $F$ is continuous on the compact set $R = [0,1] \times [0,1]$ it is uniformly continuous on $R.$ Hence there exists $\delta > 0$ such that for any pair of points $(t,u), (t',u') \in R$ with $\sqrt {(t-t')^2 + (u-u')^2} \lt \delta$ we have $$\left |F(t,u) – F(t',u') \right | \lt \frac {\epsilon} {6}.\ \ \ \ \ \ \ \ (1)$$ Now consider the partitions $$\mathcal P_1 : 0 = u_0 < u_1 < \cdots < u_m = 1\ \ \ \ \text{and}\ \ \ \ \mathcal P_2 : 0=t_0<t_1<\cdots<t_n=1$$ of $[0,1]$ such that $\left \|\mathcal P_1 \right \|, \left \|\mathcal P_2 \right \| < \delta.$ Let us define the function $\Gamma_k$ by $$\Gamma_k (t) : = F(t,u_k),\ t \in [0,1],\ k=0,1,\cdots,m.$$ Now for each $j = 0,1,2,\cdots, n,$ $\Gamma_k (t_j) \in F(R).$ Hence there exists some disk $B_s = B \left (z_s, \frac {\epsilon} {6} \right ) \in \mathcal U'$ such that $\Gamma_k (t_j) \in B_s.$ Therefore $$\left |\Gamma_k(t_j) – z_s \right | < \frac {\epsilon} {6}.$$ Now let $t' \in [t_{j},t_{j+1}].$ Since $\left \|\mathcal P_1 \right \| < \delta$ so $|t'-t_j| < \delta$ so that $\sqrt {(t'-t_j)^2 + (u_k – u_k)^2} < \delta.$ So by $(1)$ it follows that $$\left | \Gamma_k (t_j) – \Gamma_k (t') \right | \lt \frac {\epsilon} {6}.$$ Then we have \begin{align*} \left |\Gamma_k (t') – z_s \right | & \leq \left |\Gamma_k (t') – \Gamma_k(t_j) \right | + \left |\Gamma_k(t_j) – z_s \right | \\ & \lt \frac {\epsilon} {6} + \frac {\epsilon} {6} = \frac {\epsilon} {3} \lt \frac {\epsilon} {2}.\end{align*} Hence $\Gamma_k(t') \in B\left (z_s , \frac {\epsilon} {2} \right ).$ Let $D_j : = B\left (z_s, \frac {\epsilon} {2} \right ).$ Then we find that $\Gamma_k \left ([t_j,t_{j+1}] \right ) \subseteq D_j.$
Claim $:$ $\Gamma_{k+1} \left ([t_j,t_{j+1} ] \right ) \subseteq D_j.$
Since $\left \|\mathcal P_2 \right \| < \delta$ so $|u_k – u_{k+1}| < \delta$ so that $\sqrt {(t_j-t_j)^2 + (u_k – u_{k+1})^2} < \delta.$ Hence by $(1)$ we have $$\left |\Gamma_k(t_j) – \Gamma_{k+1} (t_j) \right | \lt \frac {\epsilon} {6}.$$ Now take any $t^* \in [t_j,t_{j+1}].$ So by previous discussion it follows that $$\left |\Gamma_k (t_j) – \Gamma_k (t^*) \right | \lt \frac {\epsilon} {6}\ \ \ \ \text{and}\ \ \ \ \left |\Gamma_k (t^*) – \Gamma_{k+1} (t^*) \right | < \frac {\epsilon} {6}.$$ Then we have \begin{align*} \left |\Gamma_{k+1} (t^*) – z_s \right | & \leq \left |\Gamma_{k+1} (t^*) – \Gamma_k (t^*) \right | + \left |\Gamma_k (t^*) – \Gamma_k (t_j) \right | + |\Gamma_k (t_j) – z_s| \\ & \lt \frac {\epsilon} {6} + \frac {\epsilon} {6} + \frac {\epsilon} {6} = \frac {\epsilon} {2}. \end{align*} Hence $\Gamma_{k+1} (t^*) \in B \left (z_s, \frac {\epsilon} {2} \right ) = D_j.$ Since $t^* \in [t_j,t_{j+1}]$ was arbitrarily taken so it follows that $\Gamma_{k+1} \left ([t_j,t_{j+1} ] \right ) \subseteq D_j,$ which proves our claim.
This proves that $\Gamma_k$ and $\Gamma_{k+1}$ are close together. This completes the proof.
QED
Please check my argument above. Thanks in advance.
Best Answer
Since $F$ is a homotopy defined on the compact set $R = [0,1] \times [0,1],$ $F(R)$ is compact. Since $F(R) \subseteq D$ so $F(R)$ has some positive distance from $\Bbb C \setminus D.$ Let $\text {dist} (F(R), \Bbb C \setminus D) = \epsilon > 0.$ Then for any $z \in F(R)$ we have $B \left (z, \frac {\epsilon} {2} \right ) \subseteq D.$ Consider the collection $$\mathcal U: = \left \{B \left (z, \frac {\epsilon} {6} \right )\ \bigg |\ z \in F(R) \right \}.$$ Then clearly $\mathcal U$ is an open cover of $F(R).$ Since $F(R)$ is compact $\mathcal U$ has a finite subcover. Let it be $\mathcal U',$ where $$\mathcal U' : = \left \{ B \left (z_r, \frac {\epsilon} {6} \right )\ \bigg |\ r = 1,2, \cdots, p \right \}.$$ Now let us consider the collection of disks $$\mathcal D : = \left \{ B \left (z_r, \frac {\epsilon} {2} \right )\ \bigg |\ r = 1,2, \cdots, p \right \}.$$ Since $F$ is continuous on the compact set $R = [0,1] \times [0,1]$ it is uniformly continuous on $R.$ Hence there exists $\delta > 0$ such that for any pair of points $(t,u), (t',u') \in R$ with $\sqrt {(t-t')^2 + (u-u')^2} \lt \delta$ we have $$\left |F(t,u) - F(t',u') \right | \lt \frac {\epsilon} {6}.\ \ \ \ \ \ \ \ (1)$$ Now consider the partitions $$\mathcal P_1 : 0 = u_0 < u_1 < \cdots < u_m = 1\ \ \ \ \text{and}\ \ \ \ \mathcal P_2 : 0=t_0<t_1<\cdots<t_n=1$$ of $[0,1]$ such that $\left \|\mathcal P_1 \right \|, \left \|\mathcal P_2 \right \| < \delta.$ Let us define the function $\Gamma_k$ by $$\Gamma_k (t) : = F(t,u_k),\ t \in [0,1],\ k=0,1,\cdots,m.$$ Now for each $j = 0,1,2,\cdots, n,$ $\Gamma_k (t_j) \in F(R).$ Hence there exists some disk $B_s = B \left (z_s, \frac {\epsilon} {6} \right ) \in \mathcal U'$ such that $\Gamma_k (t_j) \in B_s.$ Therefore $$\left |\Gamma_k(t_j) - z_s \right | < \frac {\epsilon} {6}.$$ Now let $t' \in [t_{j},t_{j+1}].$ Since $\left \|\mathcal P_1 \right \| < \delta$ so $|t'-t_j| < \delta$ so that $\sqrt {(t'-t_j)^2 + (u_k - u_k)^2} < \delta.$ So by $(1)$ it follows that $$\left | \Gamma_k (t_j) - \Gamma_k (t') \right | \lt \frac {\epsilon} {6}.$$ Then we have \begin{align*} \left |\Gamma_k (t') - z_s \right | & \leq \left |\Gamma_k (t') - \Gamma_k(t_j) \right | + \left |\Gamma_k(t_j) - z_s \right | \\ & \lt \frac {\epsilon} {6} + \frac {\epsilon} {6} = \frac {\epsilon} {3} \lt \frac {\epsilon} {2}.\end{align*} Hence $\Gamma_k(t') \in B\left (z_s , \frac {\epsilon} {2} \right ).$ Let $D_j : = B\left (z_s, \frac {\epsilon} {2} \right ).$ Then we find that $\Gamma_k \left ([t_j,t_{j+1}] \right ) \subseteq D_j.$
Claim $:$ $\Gamma_{k+1} \left ([t_j,t_{j+1} ] \right ) \subseteq D_j.$
Since $\left \|\mathcal P_2 \right \| < \delta$ so $|u_k - u_{k+1}| < \delta$ so that $\sqrt {(t_j-t_j)^2 + (u_k - u_{k+1})^2} < \delta.$ Hence by $(1)$ we have $$\left |\Gamma_k(t_j) - \Gamma_{k+1} (t_j) \right | \lt \frac {\epsilon} {6}.$$ Now take any $t^* \in [t_j,t_{j+1}].$ So by previous discussion it follows that $$\left |\Gamma_k (t_j) - \Gamma_k (t^*) \right | \lt \frac {\epsilon} {6}\ \ \ \ \text{and}\ \ \ \ \left |\Gamma_k (t^*) - \Gamma_{k+1} (t^*) \right | < \frac {\epsilon} {6}.$$ Then we have \begin{align*} \left |\Gamma_{k+1} (t^*) - z_s \right | & \leq \left |\Gamma_{k+1} (t^*) - \Gamma_k (t^*) \right | + \left |\Gamma_k (t^*) - \Gamma_k (t_j) \right | + |\Gamma_k (t_j) - z_s| \\ & \lt \frac {\epsilon} {6} + \frac {\epsilon} {6} + \frac {\epsilon} {6} = \frac {\epsilon} {2}. \end{align*} Hence $\Gamma_{k+1} (t^*) \in B \left (z_s, \frac {\epsilon} {2} \right ) = D_j.$ Since $t^* \in [t_j,t_{j+1}]$ was arbitrarily taken so it follows that $\Gamma_{k+1} \left ([t_j,t_{j+1} ] \right ) \subseteq D_j,$ which proves our claim.
This proves that $\Gamma_k$ and $\Gamma_{k+1}$ are close together. This completes the proof.
QED