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.


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I don’t know if this is a satisfying explanation, but the choice of radius $2\epsilon$ makes it easier to construct such disks $D_i$. Notice we only need some such covering via balls to exist and the rest of the proof is blind to what radius is used. I’m treating paths as defined on $I=[0,1]$, this is a convenience and is no different from using $[a,b]$.
Define $n:=\lfloor2/\delta\rfloor$. From now on, $k$ will be any integer between $0$ and $n$.
Define: $$\alpha_k:=\gamma_{s_1}\left(\min\left(1,\left(k+\frac{1}{2}\right)\frac{\delta}{2}\right)\right),\,z_k:=\gamma_{s_1}\left(\frac{k\delta}{2}\right),\,w_k:=\gamma_{s_2}\left(\frac{k\delta}{2}\right)$$And $z_{n+1}:=\gamma_{s_1}(1)=\gamma_{s_2}(1)=:w_{n+1}$. If
Let $D_k$ be the ball of radius $2\epsilon$ centred at $\alpha_k$. These are all contained in $\Omega$ by choice of $\epsilon$. For any $t\in I$, there is some $k$ with $k\delta/2\le t\le(k+1)\delta/2$. Then $|\gamma_{s_1}(t)-\alpha_k|<\epsilon$ by uniform continuity (notice, as in Kavi’s answer, if the change in $s$ is zero and the change in $t$ is $<\delta$, we can still make that bound) and $|\gamma_{s_2}(t)-\alpha_k|\le|\gamma_{s_2}(t)-\gamma_{s_1}(t)|+|\gamma_{s_1}(t)-\alpha_k|<2\epsilon$ (!) so both $\gamma_{s_{1,2}}(t)$ are in $D_k$.
Note $|z_k-w_k|<\epsilon$.
The $D_k$ are a covering of the two paths, then. We see, by the same bounds, that $z_k,w_k\in D_k$ for all $k$. Since the time difference between $z_k$ and $\alpha_{k+1}$ ($k<n$) is $\le 3\delta/4<\delta$, it is also true that $|z_k-\alpha_{k+1}|<\epsilon$ and $|w_k-\alpha_{k+1}|<2\epsilon$ by the triangle inequality, so $z_k,w_k\in D_{k+1}$ too - as desired.
