Set of Pairwise Disjoint Intervals is Countable

Theorem

Let $X$ be a subset of $\powerset \R$ such that:

$(1): \quad X$ is pairwise disjoint:

$\forall A, B \in X: A \ne B \implies A \cap B = \O$.

$(2): \quad$ every element of $X$ contains an open interval:

$\forall A \in X: \exists x, y \in \R: x < y \land \openint x y \subseteq A$.

Then $X$ is countable.

Proof

By Between two Real Numbers exists Rational Number:

$\forall A \in X: \exists x, y \in \R, q \in \Q: x < y \land q \in \openint x y \subseteq A$

By the Axiom of Choice define a mapping $f: X \to \Q$:

$\forall A \in X: \map f A \in A$

First it needs to be shown that $f$ is an injection by definition.

Let $A, B \in X$ such that:

$\map f A = \map f B$

By definition of $f$:

$\map f A \in A$ and $\map f B \in B$

By definition of intersection:

$\map f A \in A \cap B$

Then by definition of empty set:

$A \cap B \ne \O$

Thus by definition of pairwise disjoint:

$A = B$

Hence $f$ is an injection.

By Set is Subset of Itself, $X$ is a subset of $X$.

Thus by Cardinality of Image of Injection:

$\card X = \card {f^\to \sqbrk X}$

By definition of image:

$f^\to \sqbrk X \subseteq \Q$

By Rational Numbers are Countably Infinite:

$\Q$ is countable.

Hence by Subset of Countable Set is Countable:

$f^\to \sqbrk X$ is countable.

Thus by Set is Countable if Cardinality equals Cardinality of Countable Set the result:

$X$ is countable.

$\blacksquare$

Sources

• Mizar article TOPGEN_3:18