If $\{F_i:i\in\Bbb{N}\}$ is a collection of not empty closed sets mutually disjoint in a continuum $X$ then $X\neq\bigcup_{i\in\Bbb{N}}F_i$.

Question

What shown below is a reference from "Elementos de Topología General" by Fidel Cassarubias Segura and Ángel Tamariz Mascarúa

Lemma

If $X$ is a continuum that is covered by a sequence of numerable mutually disjoint closed sets $F_1,…,F_n$ of which at least two are not empty then for any $i\in\Bbb{N}$ there exist a continuum $C\subseteq X$ such that $C\cap F_i=\varnothing$ and at least two of the sets $C\cap F_1, C\cap F_2,…,C\cap F_n,…$ are not empty.

Theorem

If $\{F_i:i\in\Bbb{N}\}$ is a collection of not empty closed sets mutually disjoint in a continuum $X$ then $X\neq\bigcup_{i\in\Bbb{N}}F_i$.

Proof. So we suppose that the theorem if false. So $X=\bigcup_{i\in\Bbb{N}}F_i$, where $\mathcal{F}=\{F_i:i\in\Bbb{N}\}$is a partition of $X$ that is composed of closed disjoint set of which at least two are not emtpy. Since $X$ is connected, the collection of the elements of $\mathcal{F}$ that are not empty is infinite. So without loss of generality we can suppose that any $F_i$ is not empty. So for the previous lemma there exist a continuum $C_1\subseteq X$ such that $C_1\cap F_1=\varnothing$ and at least two of the sets of the sequence $C_1\cap F_\, C_1\cap F_2,…$ are not empty. So the continuum $C_1$ is equal to $\bigcup_{n\in\Bbb{N}}(C_1\cap F_i)$ and $C_1$ and $\{C_1\cap F_i:i\in\Bbb{N}\}$ respect the condictions of the previous lemma and so there exist a conitnuum $C_2\subseteq C_1$ such that $C_2\cap(C_1\cap F_2)=C_2\cap C_2\cap F_2=\varnothing$ and at least two elements of the sequence $C_2\cap F_1, C_2\cap F_2$ are not empty. So in this way we cand make a sequence $C_1\supseteq C_2\supseteq…\supseteq C_n\supseteq….$ of continuum of $X$ such that $C_i\cap F_i=\varnothing$ for any $i\in\Bbb{N}$. So this means that
$$
\Big(\bigcap_{i\in\Bbb{N}}C_i\Big)\cap\Big(\bigcup_{i\in\Bbb{N}}F_i\Big)=\varnothing
$$
and so $\bigcap_{i\in\Bbb{N}}C_i=\varnothing$. However since $X$ is compact, it must be $\bigcap_{i\in\Bbb{N}}C_i\neq\varnothing$, that is impossible.

Here for sake of completeness the text of the original proof: I hope that my was a good traslation.

So I don't understand why $\Big(\bigcap_{i\in\Bbb{N}}C_i\Big)\cap\Big(\bigcup_{i\in\Bbb{N}}F_i\Big)=\varnothing$. Indeed using induction it is possible to proof that $\Big(\bigcap_{i=1}^n C_i\Big)\cap\Big(\bigcup_{i=1}^n F_i\Big)=\varnothing$ for any $n\in\Bbb{N}$ but I don't know if $\Big(\bigcap_{i\in\Bbb{N}}C_i\Big)\cap\Big(\bigcup_{i\in\Bbb{N}}F_i\Big)=\bigcup_{n\in\Bbb{N}}\Big[\Big(\bigcap_{i=1}^n C_i\Big)\cap\Big(\bigcup_{i=1}^n F_i\Big)\Big]=\varnothing$. Moreover is it must br $\bigcap_{i\in\Bbb{N}}C_i\neq\varnothing$ because the collection of $C_i$ is a collection of closed sets of $X$ that has the finite intersection property?
Finally I don't understad why in the first part we prove that any $F_i$ is not empty if $X$ is connected: indeed for hypothesis $\{F_i:i\in\Bbb{N}\}$ is a collection of not empty closed sets.

So could someone help me please?

Answer 1

Note that for each $n\in\Bbb N$ you have

$$\left(\bigcap_{i\in\Bbb N}C_i\right)\cap\left(\bigcup_{i=1}^nF_i\right)\subseteq\left(\bigcap_{i=1}^nC_i\right)\cap\left(\bigcup_{i=1}^nF_i\right)=\varnothing\;,$$

so

$$\left(\bigcap_{i\in\Bbb N}C_i\right)\cap\left(\bigcup_{i\in\Bbb N}F_i\right)=\bigcup_{n\in\Bbb N}\left(\left(\bigcap_{i\in\Bbb N}C_i\right)\cap\left(\bigcup_{i=1}^nF_i\right)\right)=\bigcup_{n\in\Bbb N}\varnothing=\varnothing\;.$$