Proof Explanation: Cauchy's Theorem (Homotopy Version) - Why take discs of radii $3\epsilon$? I have two questions about the proof of Cauchy's Theorem (Homotopy Version) in Stein & Shakarchi's Complex Analysis. This is Theorem $5.1$, in Chapter $3$. If you have a copy, see Pg. $93-95$.


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*By uniform continuity of $F$, how do we get $\delta$ such that $$|s_1 - s_2| < \delta \implies \sup_{t\in [a,b]} |\gamma_{s_1}(t) - \gamma_{s_2}(t)| < \epsilon$$
My thoughts: 
By uniform continuity of $F$, we have for every $\epsilon > 0$, some $\delta > 0$ such that $\sqrt{(s_1-s_2)^2 + (t_1-t_2)^2}  < \delta$ implies $|F(s_1,t_1) - F(s_2,t_2)| = |\gamma_{s_1}(t_1) - \gamma_{s_2}(t_2)| < \epsilon$. To prove the required implication above, we only need to show the continuity of the map $\varphi: [0,1] \to \mathcal C([a,b], \Bbb C)$ given by $\varphi(s) = F_s$. $\mathcal C([a,b], \Bbb C)$ is endowed with the $\sup$ metric.


*What is the significance behind taking $3\epsilon$ and $2\epsilon$ in the proof? Why can't we take discs $\{D_0, \ldots, D_n\}$ of radii $\epsilon$? Pretty sure this has something to do with the claim from uniform continuity above, but I'm not able to figure it out. For $|s_1-s_2| < \delta$, we have $|\gamma_{s_1}(t) - \gamma_{s_2}(t)| < \epsilon$ for all $t\in [a,b]$, the two curves $\gamma_{s_1}$ and $\gamma_{s_2}$ could easily be contained in a union of discs of radii $\epsilon$ as well.


Reference:







$|s_1-s_2| <\delta$ implies  $\sqrt {(s_1-s_2)^{2}+(t-t)^{2} }<\delta$ for all $t$. Hence, $|F(s_1,t)-F(s_2,t)| <\epsilon$. [Just put $t_1=t_2=t$ in the inequality you got for uniform continuity of $F$].
