A set with more than $n$ components has $n+1$ pairwise separated subsets.

Question

Let $A$ be a subset of $X$ which is a compact connected metric space, and A with more than $n$ components. Do exist $C_1, \ldots, C_{n+1}$ pairwise separated subsets such that $A = \bigcup_{i=1}^{n+1}C_i$?

I have tried to take the first $n$ components of $A$ as $C_1, \ldots, C_n$ and define $C_{n+1}$ as the union of the other components of $A$ but I can't pruve that, for example, $C_1$ and $C_{n+1}$ are separated.

Answer 1

Let $M$ be the set of all numbers $m\in\mathbb N$ such that $A$ can be partitioned into $m$ disjoint nonempty sets, each clopen in $A.$

It is clear that $1\in M,$ and that $m\in M\implies\{1,2,3,\dots,m\}\subseteq M.$

If $M$ has no greatest element, then $M=\mathbb N$ and we're done. Otherwise, let $m$ be the greatest element of $M.$ Let $A=C_1\cup\cdots\cup C_m$ where the sets $C_i$ are disjoint, nonempty, and clopen in $A.$ If some $C_i$ were disconnected, then we could partition $A$ into $m+1$ disjoint nonempty clopen sets, contradicting the fact that $m$ is the greatest element of $M.$ Therefore the sets $C_1,\dots,A_m$ are connected, and they are the components of $A,$ so $m\gt n.$

Answer 2

I assume you want to require $C_i\neq \emptyset$ for all $i$ since otherwise this is trivial. It suffices to show that if $A$ is any topological space with more than $n$ components then it can be written as a union of $n+1$ disjoint nonempty clopen subsets (since then those subsets will be separated in $X$). The statement is trivial if $n=0$ so let us suppose $n>0$ and $A$ has at least $n+1$ components. In particular, $A$ is disconnected, so we can write $A=B\cup C$ where $B$ and $C$ are nonempty and clopen.

Each component of $A$ must be contained in either $B$ or $C$. Say $B$ has $b$ components and $C$ has $c$ components; then $b+c\geq n+1$. Since $B$ and $C$ are nonempty, $b>0$ and $c>0$. We can choose $b_0$ and $c_0$ such that $0<b_0\leq b$, $0<c_0\leq c$, and $b_0+c_0=n+1$. In particular, then $b_0<n+1$ and $c_0<n+1$, $B$ has more than $b_0-1$ components, and $C$ has more than $c_0-1$ components. By induction on $n$, then, we can write $B$ as a union of $b_0$ disjoint nonempty clopen sets and we can write $C$ as a union of $c_0$ disjoint nonempty clopen sets. Taking these sets together, we now have written $A$ as a union of $b_0+c_0=n+1$ disjoint nonempty clopen sets, as desired.