Cycles around compacta and the global Cauchy theorem

Question

Recall the global Cauchy theorem/formula:

Let $U$ be open and $\Gamma\subset U$ a cycle. If $\Gamma$ is homologous to zero in $U$, then for all $f:U\to\Bbb C$ holomorphic and $w\in U\setminus\Gamma$ holds $$n(\Gamma,w)f(w)=\frac1{2\pi i}\int_\Gamma\frac{f(z)}{z-w}\mathrm dz.$$

Note that "cycle" means two things in the book I'm following (Falko Lorenz's "Funktionentheorie"): initially, it was defined as a chain (finite formal $\Bbb Z$-sum) of closed integration paths (and that's how it's usually defined in formulations of the global Cauchy theorem), then the definition was expanded to mean a closed chain of integration paths, i.e. one with trivial boundary. It's not hard to see that any closed chain $\Gamma$ can be rejigged into a chain of closed paths $\Gamma'$ such that $\int_\Gamma f=\int_{\Gamma'}f$ for all $f$ (e.g. if $\Gamma=\gamma+\overline\gamma$, where $\overline\gamma$ is the reverse path, then we might take $\Gamma'=\gamma*\overline\gamma$), so it doesn't really matter which definition is used.

I am trying to understand the proof of the following proposition

Theorem: Let $U$ be open and $K\subset U$ compact and non-empty. There exists a cycle $\Gamma\subset U\setminus K$ such that for all $f:U\to\Bbb C$ hoolomorphic and $w\in K$ holds $$f(w)=\frac1{2\pi i}\int_\Gamma\frac{f(z)}{z-w}\mathrm dz,$$ i.e. there is a cycle in $U\setminus K$ such that the Cauchy formula holds for it on $K$.

This is a weaker version of the statement that we can always find a cycle $\Gamma\subset U\setminus K$ that is homologous to zero in $U$ and winds around $K$ once (a proof can be found e.g. here, on page 160).

The proof of the above theorem runs roughly as follows

Pave the plane with $\varepsilon(\Bbb Z+i\Bbb Z)$ and let $Q_1,\dots,Q_n$ be the finitely many squares of the lattice intersecting $K$; by making $\varepsilon$ small enough we can ensure that $Q_i$ are all contained in $U$. Let $$\delta_i=\gamma_{i1}+\gamma_{i2}+\gamma_{i3}+\gamma_{i4}$$ be the positively oriented boundary of $Q_i$, let $\Lambda=\sum_i\delta_i$, and let $\Gamma$ be the sum of those sides $\gamma_j$ that belong to only one $Q_i$, then $\Gamma$ is a cycle in $U\setminus K$ and $\int_\Gamma f=\int_\Lambda f$ for all $f\in H(U)$. We then show that the formula holds for $z\in K$ not lying on $\Lambda$ and extend it to all of $K$ by continuity, since both sides of the identity are analytic.

I am trying to understand the highlighted part, namely why $\Gamma$ is a cycle (in either sense of the word) and why it must be contained in $U\setminus K$ (if a side of $Q_i$ does not belong to any other square, why should it not intersect $K$?). Both statements make intuitive sense to me, however I can't prove them formally.

If there are alternative proofs/expositions of this theorem or its aforementioned stronger version, I'd be interested in seeing them.

Addendum: I have since found out that this result appears in most books that prove Runge's theorem (a notable example being Rudin's "Real & Complex Analysis", Theorem 13.5), with pretty much exactly the same proof.

Addendum: To the people saying the summary of the proof I posted must be wrong, I'm posting the proof from Rudin's RCA, which is exactly the same.

Answer 1

The argument is sloppy since it allows $\Gamma$ to intersect $K$. A clean way to argue is as follows.

1. Define $\eta$ as in Rudin's paper (i.e. every point of $K$ is at distance $\ge 2\eta$ from the complement to $\Omega$) and $\epsilon=\eta/3$. Let $K_\epsilon$ denote the closed $\epsilon$-neighborhood of $K$, i.e. the set of points $x$ such that the minimal distance from $x$ to $K$ is $\le \epsilon$. Then $K_\epsilon\subset \Omega$ ($\Omega=U$ in your notation) and $K$ is contained in the interior of $K_\epsilon$. Then argue almost as in Rudin and consider those squares with edge-size $\epsilon$ which intersect $K_\epsilon$. The union $W$ of these squares is still contained in $\Omega$ and $K$ is contained in the interior of $W$. To see that $W\subset \Omega$ note that every point of $W$ is within distance at most $2\sqrt{2}\epsilon$ from a point in $K$. But $2\sqrt{2}\epsilon< 2\eta$ since $\epsilon=\eta/3$.

2. Then form the 2-dimensional chain $$ C=\sum_{j} Q_j $$ summing up all the squares used above with the orientation induced from the complex plane. Consider the (algebraic) boundary $\partial C$ of this chain: $$ \partial C= \sum_j \partial Q_j $$ where each $\partial Q_j$ is what you denoted $\delta_j$. This is your 1-cycle $\Gamma$. The key fact is that $$ \partial (\partial c)=0 $$ for each 2-chain. Since $\partial$ is a linear operator, to verify this identity you just check it for each square $Q$, where it is manifestly true: $$ c= \sum_j a_j Q_j, a_j\in \mathbb Z, $$ $$ \partial c= \sum_j a_j \partial Q_j, $$ $$ \partial \partial c= \sum_j a_j \partial \partial Q_j=0. $$ (This is a baby version of what you see in an algebraic topology class if you ever take one.) Thus, $\partial \Gamma=0$. The fact that $\Gamma$ is a cycle in $\Omega$ follows from the fact that $W\subset \Omega$ and, hence, every edge of a square in $W$ is also contained in $\Omega$. The fact that $\Gamma$ is disjoint from $K$ follows from the fact that $\Gamma$ is an algebraic sum of edges each of which is contained in the topological boundary of $W$ and the latter is disjoint from $K$ by the construction.

Answer 2

Every edge of the lattice belonging to one of the $Q_i$ either belongs to two squares with opposite orientation, or to only one square in which case it belongs to $\Gamma.$ This means that $\Gamma$ is actually equal to a sum of cycles (squares and two-cycles consisting of the sum of twice the same edge with opposite orientation), and therefore a cycle.

You are right about the argument being incomplete as regards the statement that $\Gamma$ does not intersect $K,$ but it can be completed without further decreasing $\epsilon.$ The trick is that if a side of a square intersects $K,$ so do the (closed) squares on both sides and therefore that segment does not belong to $\Gamma.$