Cauchy's integral theorem requirement of open set

Question

Cauchy's integral theorem states the following: Let $ U\subset{C} $ be an open star domain, $ f:U\rightarrow{C} $ a function that is analytic on U, and $ \gamma:[0,1]\rightarrow{U} $ a partwise continuous and closed path. Then $∫_\gamma f(z)dz=0$.

I can't see exactly why the condition of $U$ being open is vital for the theorem to hold.

We proved the theorem in class through Green's theorem and also specifically for triangular paths, and when going over the proof for triangular paths I couldn't find where we used the openness of $U$.

Any intuition on this would be greatly appreciated..

Answer 1

A definition of an analytic function demands that the domain be open, otherwise the condition of differentiability (or expandability into a power series) wouldn't make sense. So the requirement is not specific to the Cauchy's theorem.

It is conceivable that one could define restricted differentiability (and in a similar fashion - expandability into a power series) on a set $U \subseteq \mathbb{C}$ that is not open. To do that, we demand that for each $z_0 \in U$ the fraction $\frac{f(z)-f(z_0)}{z-z_0}$ converges to a complex number (which we name $f'(z_0)$) as $z \to z_0$ goes only through the elements of the domain $U \setminus \{ z_0 \}$. But then

• If $z_0 \in U$ is isolated, then every complex number would be a valid limit of $\frac{f(z)-f(z_0)}{z-z_0}$ as $z \to z_0$ restricted to $z \in U \setminus \{ z_0 \}$;
• Doing usual analysis would be harder;
• I don't know of any decent advantages of such approach over the usual one.