Problem 1555

To show that any two intervals $A, B \subseteq \mathbb{R}$ of positive lengths can be countably dissected into each other, we need to demonstrate that they can be written as countable unions $A = A_1 \cup A_2 \cup \ldots$ and $B = B_1 \cup B_2 \cup \ldots$ of pairwise disjoint sets, where $A_i$ and $B_i$ are congruent for every $i \in \mathbb{N}$.

1. **Case 1: Open Intervals $A = (a, b)$ and $B = (a', b')$**

- Define $a_0 = a$ and recursively define $a_{n+1} = \frac{a_n + b}{2}$. This creates a sequence of nested intervals:
$$A = \bigcup_{n \in \omega} (a_n, a_{n+1}] \setminus \{b\}$$
- Similarly, define $a'_0 = a'$ and recursively define $a'_{n+1} = \frac{a'_n + b'}{2}$:
$$B = \bigcup_{n \in \omega} (a'_n, a'_{n+1}] \setminus \{b'\}$$
- Note that $b \notin \bigcup_{n \in \omega} (a_n, a_{n+1}]$ and $b' \notin \bigcup_{n \in \omega} (a'_n, a'_{n+1}]$.

- Define $I(n) = (a_n, a_{n+1}]$ and $I'(n) = (a'_n, a'_{n+1}]$. Each $I(n)$ is congruent to $I'(n)$ because they are intervals of the same length.

2. Case 2: Other Types of Intervals

- For intervals that include boundary points, such as $A = [a, b]$, $A = [a, b)$, or $A = (a, b]$, we can include the boundary points in the first interval $I(1)$ and proceed similarly to Case 1.

- For example, if $A = [a, b]$, we can write:
$$A = \{a\} \cup \bigcup_{n \in \omega} (a_n, a_{n+1}] \cup \{b\}$$
- Similarly, for $B$, include the boundary points in $I'(1)$.

3. Ensuring Pairwise Disjoint Sets

- It is important to show that $\forall m, n$, $I(m) \cap I(n) = \emptyset \iff m \neq n$. This ensures that the sets $A_i$ and $B_i$ are pairwise disjoint.

4. Cardinality Consideration

- Any nonempty subinterval of $\mathbb{R}$ has the cardinality of the continuum. This can be shown using the Schröder-Bernstein theorem, which states that if there exist injective functions $f: A \to B$ and $g: B \to A$, then there exists a bijection between $A$ and $B$.

By following these steps, we can construct the required subsets $A_1, A_2, \ldots$ of $A$ and $B_1, B_2, \ldots$ of $B$ such that $A_i$ and $B_i$ are congruent for every $i \in \mathbb{N}$.

$\blacksquare$