Elementary topology of $\mathbb C$: Union of 2 regions with nonempty intersection is a region

Question

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.31,32

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Perhaps related but my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis: union-of-connected-subsets-is-connected-if-intersection-is-nonempty, Union of connected subsets is connected if intersection is nonempty

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Pf:

Let $A$ and $B$ be the regions in question.

1. Their union is open is because they are each open. If that warrants further justification, then: any element in their union has a disc centred at the element that is contained entirely in A or in B and hence in A or B.

2. Suppose on the contrary that their union is not connected. Then $\exists C_1, C_2, C_3, C_4$ s.t. $C_1 \cup C_2 = A \cup B, C_1 \ne \emptyset \ne C_2, C_3 = C_3^0, C_4=C_4^0, C_1 \subseteq C_3, C_2 \subseteq C_4$ and $C_3 \cap C_4 = \emptyset$.

We deduce that $\emptyset = C_1 \cap C_2$, and so we have 4 cases:

Case 1: $A \cup B \subseteq C_1$

$A \cup B \subset$ but $\ne C_1 \cup C_2$ ↯

Case 2: $A \cup B \subseteq C_2$

$A \cup B \subset$ but $\ne C_1 \cup C_2$ ↯

Case 3: $A \subseteq C_1, B \subseteq C_2$

$A \cap B = \emptyset$ ↯

Case 4: $A \subseteq C_1, B \subseteq C_2$

$A \cap B = \emptyset$ ↯

The cases are exhaustive and each lead to a contradiction. $\therefore,$ their union is connected. QED

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For the former my proof is:

Definitions:

$$z \in \partial A \iff \exists \ r>0:D[z,r]\cap A \ne \emptyset \ne D[z,r]\cap A^c$$

$$z \in \partial B \iff \exists \ r>0:D[z,r]\cap B \ne \emptyset \ne D[z,r]\cap B^c$$

$$B \ \text{is closed} \iff \partial B \subseteq B \iff B = B^0 \cup \partial B$$

Now observe for $D[z,r]\cap A$:

$$D[z,r]\cap A \subseteq A \subseteq B \implies D[z,r]\cap B \ne \emptyset$$

Thus for $D[z,r]\cap A^c$:

$$D[z,r]\cap A^c \subseteq B \iff D[z,r] \subseteq B \iff z \in B^0$$

$$D[z,r]\cap A^c \subsetneq B \iff D[z,r]\cap B^c \ne \emptyset \iff z \in \partial B$$

In either case $z \in B$. QED

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For the latter my proof is:

$$\forall z \in A=A^0, \exists \ a>0: D[z,a] \subset A =A^0 \subseteq B$$

Now $$z \in B^0 \iff \exists \ b>0:D[z,b] \subseteq B$$

Choose $b=a$. QED

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Where have I gone wrong?

Answer 1

In any space let $F$ be a non-empty family of connected open sets such that $\cap F\ne \emptyset.$ Let $A, B$ be disjoint open sets with $A\cup B=\cup F.$

For each $f\in F,$ the sets $f\cap A , f\cap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...\quad f\cap A\ne \emptyset \implies f=f\cap A \quad \text { and } \quad f\cap B\ne \emptyset \implies f=f\cap B.$$ By contradiction, suppose $A\ne \emptyset \ne B.$ Since $A=\cup_{f\in F}(f\cap A)$ and $B=\cup_{f\in F}(f\cap B),$ there exist $f_A,f_B\in F$ such that $f_A\cap A\ne \emptyset \ne f_B\cap B.$

Then by $(1)$ we have $\cap F\subset f_A\cap f_B=(f_A\cap A)\cap (f_B\cap B)\subset A\cap B=\emptyset,$ contrary to $\emptyset \ne \cap F.$

Answer 2

Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete {0,1}, f is constant.

If C is a collection of connected sets, a in $\cap$C,
and f:$\cup$C -> {0,1} continuous, then for all K in C,
f(K) = {f(a)}. Whereupon, f($\cup$C) = {f(a)}.