component of the union of two sets This question is exercise 2 on page 32 of the book Analytic Functions by Stanislaw Saks and Antoni Zygmund:
(All sets are in the extended complex plane) If S, disjoint from a certain closed set Q, is a component of the closed set P, then S is also a component of the set $P\cup Q$.
I came up with a possibly wrong idea for a proof using exercise 1 on the same page (If P and Q are closed sets, then the union of those components of the set P which have points in common with Q is also a closed set): Let C be the component of $P\cup Q$ which contains S, B the union of all components of P which contain points of Q, and A the union of all components of P which don't contain points of Q but are contained in C. Then $S\subset A$ and $C=A\cup ((B\cup Q)\cap C)$. Assuming S is properly contained in A you can write A as the union of two disjoint nonempty sets closed in A (call them $A_1$ and $A_2$) one of which contains S because S is a component of P. If A were closed in C (or contained in a set closed in C and disjoint from $((B\cup Q)\cap C)$) you would then have a decomposition of C into two nonempty disjoint sets closed in C (namely, $A_1$ and $A_2\cup ((B\cup Q)\cap C)$). But I couldn't prove that A was closed in C or contained in a larger set closed in C and disjoint from the other set.
 A: The key topological fact is that the extended complex plane (the sphere) is a compact Haussdorf space, which we denote $X$. Then $P$ is also compact Hausdorff, and $S$ is a quasicomponent: $S=\bigcap_E E$ where the intersections runs in all clopen subsets $E$ of $P$ containing $S$ (see Quasicomponents and components in compact Hausdorff space); in particular, the complements $P\setminus E$ are also clopen in $P$. Note also that $S$ is closed in $P$, which is in $X$, hence $S$ is closed in $X$. We argue as follows to show that $S=C$.
1) If $C\cap P=S$, then $C$ is a connected component of $S\cup Q\subset P\cup Q$ and since $S$ and $Q$ are closed and disjoint, $C$ must be contained in one of them, hence coincides with $S$ as wanted.
2) Suppose there is $x\in C\cap P\setminus S$. Then $K=\{x\}\cup(P\cap Q)$ is a closed set disjoint from $S$, so that $K\subset \bigcup_E(P\setminus E)$ (the $E$'s above). Since $K$, being closed in $X$, is compact, there are finitely many $E_1,\dots,E_r$ with
$K\subset\bigcup_i(P\setminus E_i)=P\setminus\bigcap_iE_i$. We set 
$F=\bigcap_i E_i$ is clopen in $P$ as all $E_i$ are, hence $F$ and $P\setminus F$ are closed in $P$, hence in $X$. We write:
$$
P\cup Q=F\cup\Big((P\setminus F)\cup Q\Big).
$$
The right hand is a disjoint union of two closed sets: $F$ and the big parenthesis. Indeed, notice that $F\cap Q\subset F\cap P\cap Q\subset F\cap K=\varnothing$. Consequently, since $C$ is a connected component of the union, cannot intersect both disjoint closed sets. However, $S\subset F$ is contained in $C$, and $x\in C\cap K\subset P\setminus F$. This contradiction completes the argument.
