Exercise 1.5.6

Proof.

Here we write an open interval $]a,b[$, rather than $(a,b)$.

b) We will prove by contradiction that no uncountable collection of disjoint open intervals exists.

Assume that ${(J_i)}_{i\in I}$ is a family of disjoint open (non empty) intervals $J_i=]a_i,b_i[$ $(a_i<b_i)$, where $I$ is uncountable.

$\text{∙}$

Reduction to a bounded interval.

Consider, for each $n\in \mathbb{N}$,

$$I_n=\{i\in I\mid J_i\subset ]-n,n[\}$$

Since every $J_i$ is included in $]-n,n[$ for some naturel number $n\in \mathbb{N}$, we obtain

$$\underset{n\in \mathbb{N}}{\bigcup }{I}_n=I.$$

Indeed $I_n\subset I$ for all $n\in \mathbb{N}$, so ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \mathbb{N}}{I}_n\subseteq I$, and conversely, if $i\in I$, then $J_i\subset ]-n_0,n_0[$ for some $n_0\in \mathbb{N}$, so $i\in I_{n_0}\subseteq {\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \mathbb{N}}{I}_n$.

If every $I_n$ was finite or countable, then $I={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n\in \mathbb{N}}{I}_n$ would be finite or countable (Theorem 1.5.8(ii)), and this is a contradiction, since $I$ is uncountable by hypothesis.

This shows that there is an uncountable set $K=I_n$ of indices such that the family ${(J_i)}_{i\in K}$ is a family of disjoint open intervals $J_i=]a_i,b_i[$, where $J_i\subset ]-n,n[$ for every $i\in K$.

$\text{∙}$

Measure of the union.

Write $l_i=l(J_i)=b_i-a_i$ the length of the interval $J_i$.

Consider now, for each $m\in \mathbb{N}$,

$$K_m=I_{n,m}=\{i\in K\mid l(J_i)>1∕m\}.$$

Since $J_i\ne \varnothing$, for all $i\in K$, there is some $m\in \mathbb{N}$ such that $l(J_i)>1∕m$. Hence

$$\underset{i\in K}{\bigcup }{K}_m=K.$$

The argument given in the first part shows that there is some $m\in \mathbb{N}$ such that $K_m$ is uncountable. a fortiori infinite: there are indices $i_1,i_2\dots ,i_{3\mathit{nm}}$ in $K_m$ such that

$$J_{i_1}\cup J_{i_2}\cup \dots \cup J_{i_{3\mathit{nm}}}\subseteq ]-n,n[.$$

Since the $J_i$ are disjoint,

$$l(J_{i_1})+l(J_{i_2})+\cdots +l(J_{i_{3\mathit{nm}}})\le l(]-n,n[)=2n.$$

By definition of $K_m$, $l(J_{i_k})>1∕m$ for every $k\in K_m$. Therefore

$$3n<2n.$$

Thus $3<2$. This is a contradiction. Therefore, no uncountable collection of disjoint open intervals exists.